Transcript for:
Understanding Complex Numbers Basics

Hello kids, welcome to Science and Fun where teachers teach you with both heart and mind. So today we are not going to study complex numbers one shot, today we will study super one shot of complex numbers. Super one shot because you will not need to watch scattered lectures. Although a few months ago we have uploaded one shot of complex numbers on this channel. A lot of love has been given to the kids there. But inside that there were only and only NCRT questions. After that we made another lecture where there were some important questions But this time we thought why not we give you such one shots And all my one shots are going to come from the front Where we give you a whole package Means in one shot you will get the entire solution of NCRT All the concept clearing will be given That concept which will help in extra questions And those PYQs will be given in this lecture Those most expected questions will be given in this lecture You can say NCRT, exemplar and questions from children's schools or the questions that I think, they didn't come but I have full hope of them coming. If you have read the complex number chapters in school, for example, you must have heard that there are questions with locus, straight lines, purely imaginary, purely real. So we are going to do some good MCQs and some good subjective questions here. So, this chapter is not going to be just about NCRT. Especially this chapter. And there will be some extra questions for each chapter. Because MCQs are not in our NCRT. So, the first chapter of core mathematics that I am teaching you is complex number. This is the fourth chapter of NCRT. Why didn't I start the sets chapter from there? Trigger the sets relation function. Because I thought that sets and relation function must have been done in the schools of children. We will do that later on as a revision mode. But we will start with the complex number first. And in this lecture, you will get a whole wholesome package. But you will have to take a pen copy. You will not be able to study without it. And in between, your brother will keep guiding you that now you have to take a break. Or now you have to do this and come back to the lecture. Is it clear? So all the questions, that is, you don't need to see anything else till the finals. Just make this lecture your friend. And it will continue like this. And all my lectures are going to be like this. Which lecture will be next, you can also tell me in the comment section. So that I can start its preparation in this way. Because it takes research. Because if you go to the end and see the question, you will feel that yes. And how are the questions going? First concept, then question, concept, question, concept, question. And the questions are going to be of NCRT and extra as well. Along with that. And last but not least, there are 10-12 questions which are very important. Come on, we will start without wasting any time. So what is a complex number? We have been teaching about a number since childhood That is real numbers So in 9th standard there was a chapter on number system In 10th standard there was a chapter on real numbers So there we had read a thing called real numbers Which was said to be rational and irrational Now what is a complex number? Okay What is a complex number? Complex numbers is Combination of Combination of of real numbers and imaginary numbers. And imaginary numbers. Now, we have complete knowledge about real numbers. We have been studying them for many years. Now, what are these imaginary numbers? Those numbers which do not exist in reality, they are just a part of our imagination. Now, what are such numbers? which is only a part of our imagination So these numbers are called Iota Why is it imaginary? Because the value of this is imaginary which is the under root of minus 1 which is a sin in reality We have never seen a negative under root For example, I want to take you on a flashback Earlier this chapter was called complex numbers and quadratic equation. Now the quadratic equation we had studied in 10th also. So why were we studying quadratic in 11th? Let me tell you. For example, you must know the formula of d. d was b square minus 4ac. If d was greater than or equal to 0, then our real roots would come. Real and distinct or real and equal. But if our d... If less than zero was present, we would say that there is no real root. In 10th standard, we would say that there is no real root present. But nobody would tell you that there is no real root, but imaginary root is present. An offline student asked, why are we studying the imaginary? Why are we studying the imagination? If we are being taught maths, it is for engineering When we are being taught maths, we don't have to study the imagination We will have to study the reality and calculate There are derivatives and integration It helps a lot in our engineering process I am saying the right thing So I believe, or what my teachers used to tell me, that the topics of maths that have been invented, the topics of maths that have been invented, are due to physics, due to chemistry, due to science. Maximum topics have been invented because of physics. So I believe, or my teachers used to tell us, that they believe that when a mathematician or a scientist was inventing something, or when an equation was being invented, were looking for a solution, a process for science they must have been stuck somewhere for example, they must have thought that somewhere x2 is equal to minus 4 now this is not present in the real world but they have to move forward so they made a rule of their own they said that it doesn't exist in reality but if this minus is made to disappear if this minus is made to disappear for example, here the value of x2 was minus 4 So the value of x is under root minus 4. If I remove minus, then it will be a problem. So what did they say? Under root minus 1, we take iota. So now our answer is 2 iota. Where iota is my imagination. Do you understand? This story will be there. I believe so. Teachers believe so. Or their teachers believe so. Okay? So here, iota is just an imaginary number. Since the birth of iota, the imaginary number, they have made a whole area, a whole universe, of complex numbers. And we have to study that. Now, the world we are going to study, of complex numbers, is just an imagination, which solves those questions, cracks those equations, which doesn't exist in reality, but yes, you have to solve it you can solve it by taking an iota so the story is this, it's just an imagination and the first chapter of this, complex numbers and quadratic equations so here iota was used back to the questions, back to the concept imaginary numbers are iota i means iota iota is under root minus 1 Now what does under root minus 1 mean here, dear children? You can also remember it like this, that the value of iota, the value of iota square, what is that? Minus 1. You just have to remember this, that the value of iota square here will be minus 1. Is it clear? So one is the real number, one is our imaginary number, and the value of iota square here will be minus 1. Now, the components of our... Complex numbers have a standard form. Standard form. That is Z. Z means complex number. We represent complex number with Z. Now you say, Sir, Z is an integer. It is capital Z. It is small Z. Z means A plus iota B. Now A stands for real part. What is this, my son? Real part. And B stands for... imaginary part. Okay? If I tell you, the 98% content of the whole chapter is this, brother. We have to find out anything in the whole chapter. First of all, where do we have to go? We have to go to the standard form. If you reach here, then the way forward is very easy. So, first of all, we are going to practice the standard form. Okay? Various types of questions. There are many questions in the paper to convert into standard form. So Z, or complex number, is real plus imaginary part. The adjacent coefficient of iota will be called imaginary part. And the constant will be called real part. Real means A can be rational or irrational. B can be rational or irrational. But if it multiplies with iota, Holy, what is that? It is an imaginary part. If we add these two, then what is this? A complex number. It's so simple, there is no complex number There is addition of real part and imaginary part There is combination, we call it complex number Remember the value of iota, under root minus 1 That is, remember the value of iota square, minus 1 It's so simple Now the whole world has started When you studied real world, you studied graphs You studied many things, here also many graphs and other things will come But first, we make this thing our friend Because 98% of the content is the standard form. Like this. Now we will come back to this. Here is an empty page. And in this, we will write some of our formulas. Okay? Let's go ahead. So now, like this is a question. It is a very simple question. Now, here, see one formula. It came here. Like here. A point to remember. We will make this page. We made it in the starting. Point to remember. Remember. Oh, brother. Remember. And here, what are you doing? Point to remember. Like what is the first point? If two complex numbers are equal, then real part is equal to real and imaginary part is equal to imaginary. If z1 is equal to z2, that is, a plus iota b, if it is equal to x plus iota y, then therefore, a will be equal to x. And b will be equal to y. Okay? a will be equal to x and b will be equal to y. We will keep this in mind and apply it in the question. What is it saying? It is saying that 4x plus 3x minus y iota is equal to 3 plus minus 6 iota. It is saying this. These two complex numbers are given equal. Find the value of x and y. So, we will write here, kids, on comparing. Real part will be equal to real. It is a very simple question. 4x will be equal to 3. And 3x minus y will be equal to minus 6. Now, you can solve this. The value of x is 3 by 4. And if I put the value of x here, 3 by 4. So the value of y will crack here. Okay? This here, this here. So 9 by 4 plus 6 is equal to y. 9 plus 24 by 4 is equal to y. So the value of y is 33 by 4. This is my answer. Okay? Go to the process. Calculations are for kids. What are you saying? If two complex numbers are equal, then the real part will be equal to the real, and the imaginary part will be equal to the imaginary. The conclusion that will come out from there will be your answer. First, concept understood now many more questions 10-12 questions will only indicate one thing you have to convert in standard form a plus b iota or a plus iota now normal mathematics will be used look what is written minus 5 iota and 1 by 8 iota Don't forget the normal mathematics, LCM, SCM, factorization, rationalization, all of that is basic. We will solve it. What is it? Multiply it. Minus 5 Iota square divided by 8. Is it clear? So what is the value of Iota square? Minus 1 divided by 8. So what is our answer? 5 divided by 8. So when we write our complex number, what is the standard form? It is A plus Iota B. Now in a plus iota b, we have only real part. Only a. And there is no iota b. So what will we write? Zero iota. We will write this. Because to write only complex numbers, we need real part and imaginary part. Here real part is not imaginary. It is not. So what is it? It is zero. And see, as soon as this question came, minus iota, 2 iota, minus 1 by 8 iota cube. So you can multiply these. So this is minus 2 iota square. Okay. Now minus 1 by a iota cube. So this is done. Minus iota cube. And 8 cube is 512. This is done. Okay. Now understand. Iota. Now understand. I am going to tell you a very important thing. What is minus minus? Plus. Cut this 512 from this 2. Okay, this is 256. Now, Iota's power is 5 upon 256. This is our answer. Now let's come to point to remember again. Now see, whenever the power of iota comes, bigger than 2, bigger than 2, so let's go back to point to remember. Second point to remember, suppose wherever it is written iota cube, then it will have to be broken. You know the value of iota square, you know the value of iota power 1, but to get the value of iota cube, what will you have to do? iota into iota square. iota into iota square. See, base is same. Power is added, so 3 will come. Now, iota is iota. What is this? Minus 1. So, the answer is minus iota. So, iota cube's value is minus iota. Wait, wait. Somewhere it is written, iota's power is 4. Now, this is even. If it is even, then it will break directly. See, iota square's square. Okay? Now, here comes minus 1's square. That means, the answer is 1. Somewhere it is written that Iota has power 5. Try to understand. This page will help a lot in the future. Then here comes Iota into Iota's power 4. Iota is Iota. And this becomes Iota square. Oh, it has become a very big square. Iota square. What square? Now Iota is Iota. Minus 1 square which is 1. The answer is Iota. Somewhere it is written that Iota has power 16. So this is even Even will break like this Square will become power Now if minus 1 is 8 Then answer is 1 If Iota is 39 Then we will have to break it Iota into Iota is 38 Now 38 is even power So what did we write? Iota square is 19 Now Iota minus 1 is 19 So the answer is Yep minus iota. Right? Answer iota, minus iota, 1 or minus 1. So, this way, we have to break it. See, iota's power is 5. What can I write? iota into iota's power 4. Upon, I got 256. Right? Now, iota's power 4, what happened? iota square. What square? Square is getting bigger. upon is 256 what is the square of minus 1? 1 so the answer is iota upon 256 but if we want to do it in standard form then it is only imaginary part not real part so real part is 0 plus iota upon 256 this is our answer so slowly you are learning and moving forward so the power of iota is base same power is added so it is 5 so the power of iota is There are many more questions. There will be some weird questions like 35, 9, 79, etc. I am teaching you all these. Don't worry. This topic will come again. So, you have to do normal mathematics and follow the new rules. Now, what is this? Minus under root 3. Did you understand? But what is minus under root? Which is a sin. And this sin can only solve an imaginary number. Come. So what can we do? Under root 2, instead of minus 1, we have iota. Right? I told you this formula. And here comes 2 root 3 minus iota. Multiply it. It's normal mathematics. If you multiply it with this, then it will be minus 6. If you multiply it with this, plus under root 3 iota, if you multiply it with this, then it will be plus 2. root 6 iota will come. If you multiply this by this, then minus root 2 iota square will come. Okay? So minus 6 plus under root 3 iota plus 2 root 6 iota and along with this plus under root 2. So real ones will always come in one bracket. And imaginary ones will come in one bracket. This is what we got. Okay? So this is our z. Real plus imaginary. Right? Is it clear? Now, we have only one focus. To convert it into standard form. Let's see. Now, as soon as this question comes. You know. See, what is standard form? First tell me. A plus iota B. Is iota in numerator or denominator? Iota is in numerator. So, first of all, if iota is in denominator, then we have to remove it. How? Rationalization is the only tool. If iota is in denominator. So we have to remove it from rationalization. Point to remember. So this page will be a great one for you. If... denominator denominator in iota rationalization ok do it its your copy write it in any language we have to make notes the language we understand is good ok so now under root So, it is below. We had read in our childhood that if we want to remove under root, then rationalization. And if we want to remove iota, then also rationalization. Here, we have 5 plus under root 2 iota. 1 minus under root 2 iota. We will rationalize here. This came like this and this came like this. And I hope that you will know rationalization. Do it, brother. 5. Do it here, brother. 5 under root 2 iota. You will do this under root 2 iota. you will do this 2 iota square divided by a minus b, a plus b so a square minus b square 5, 5 root 2, 1 root 2 6 root 2 iota now see, iota square's value is minus 1 so what will happen? minus 2 we will keep this in mind 1, and this is 2 iota square 2 out of 5, how much will come? 3 plus 6 under root 2 iota divided by 1 plus 3. 1 plus 2, sorry. What is the value of iota square? Minus 1. 1 plus 2 is done. So, here it is. 3 plus 6 root 2 iota upon 3. You can do it differently. 3 by 3. 6 root 2 by 3 iota. The answer is 1 plus 2 root 2 iota. And what is this? Our standard form. Okay. Go with the flow. You have to keep one thing in mind. iota square's value is minus 1 rest normal mathematics apply as much as you can ok see more don't worry iota's power is minus 35 no tension iota's power is minus 35 iota's power is minus 35 see first of all iota should not be below and when iota's power increases how to crack I have taught you how The power of iota is odd. So iota into iota's power is 34. iota, here it comes, iota squared's power is 17. iota minus 1's power is 17. Which is minus 1. So till now the answer is 1 upon minus iota. And in denominator we don't leave iota. So therefore what we have to do is rationalization. So, into iota, divide by iota. I hope you remember. To rationalize 1 by root 2, we used to multiply by root 2 and divide by root 2. So, what will you do to do 1 by iota? Into iota, divide by iota. So, minus iota upon iota square. Minus iota upon minus 1. So, the answer is iota. Now, what is the standard form of iota? It is not the real part. So 0 plus iota is my answer. It's such a simple question. Maximum things are in standard form. We have converted iota into standard form. Till now, Phil Hall. Phil Hall, not Phil Hall. So we have to convert it into standard form. Iota's power is 9. Iota's power is 19. So see, 9. And here comes 19. What can I take from iota's power 9? Iota into iota's power 8. Because it was even. Here, iota into iota's power is 18. iota, iota square's power is 4. iota, iota square's power is 9. iota square's power, what is square? Minus 1. And here also, what is minus 1? Minus 1's power is 4, 1. And minus 1's power is 9, minus 1. iota minus iota, the answer is 0. So our standard form is 0 plus 0 iota. It must have become flow, it must be fun. I am really enjoying it. Okay? It has become flow, it is fun. Look more. Look more. Hey, brother, give me such a big power. It is not a matter of taking tension. No. Iota's power is 75. Take 75. Okay? So look. Here comes iota's power 75. Iota's power 80. Iota's power is 85, Iota's power is 90. Sir, can we take iota's power 75 as common? Yes, we can. Take it like this. iota's power 75. Or solve it separately as we did. Here comes 1. Here comes iota's power 4. Not 4, 5. Here comes iota's power 10. And here comes iota's power 15. We have to do that work. You cannot escape from what you think you will escape from. You have to do that work. See here also, we have to do it 4 times only. So it's better not to get into this mess. What to do? Solve it straight away. Here comes Iota into Iota's power 74. Here comes Iota squared's power 40. Here comes Iota into Iota's power 84. Here comes Iota squared's power 45. Even ones break directly, and odd ones have to break. Here comes Iota squared's power 37. this is minus 1 to the power 40 this is iota square to the power 42 minus 1 to the power 45 now here minus 1 to the power 37 this is 1 here minus 1 to the power 42 and this is minus 1 so here minus iota plus 1 plus iota minus 1, everything will be cut, the answer is 0 plus 0 iota now you must be feeling, I mean you don't feel the need sir move aside, we will do the question we will do it, now whatever question comes next, you will pause the video and do it yourself if you follow my instructions then this chapter will be clear like a mirror this is my promise to you just follow this instruction now try this question yourself, just once, no problem, okay? and then, till the standard form is running, you will try it yourself and then you will turn on the video very good yellow color is best now here comes Iota's power n Iota's power n plus 1 Iota's power n plus 2 and here comes Iota's power n plus 3 so out of all, Iota's power n is common now why is it important to take common here? see So, it is not necessary to take common. If you think that taking common is mandatory, the answer is no. Without that, it will also be possible. Look, Iota power n. This is Iota power n into Iota. This is Iota power n into Iota square. This is Iota power n into Iota cube. Okay, base same power add. This will remain same. This is also same for now because Iota power is 1. Now here comes our minus 1. And now just remember what iota cube is. Minus iota. Okay. Otherwise the method is the same. Let me do it one more time. Last. iota into iota square. Okay. So here comes iota's power n. iota's power n iota. Minus iota's power n. What is this? Minus 1. So minus iota's power n into iota. Everything is cut. See nothing is left. Everything is cut. Answer is 0 plus 0 iota. Right kids? This is the answer. Zero is the answer. Zero or zero plus zero iota. And look at this question. A plus B has a whole cube. So you must know its identity. I hope I am sitting with this in mind. That we will know what is A plus B's whole cube. Is this visible? So this is A cube plus B cube plus 3A square B and 3AB square. Okay? And if it was minus, then here also minus, here also minus, here also minus. Right? No. a cube minus b cube minus 3a square b plus 3ab square. See that. The formula is this. a minus b whole cube. a cube minus b cube minus 3a square b plus 3ab square. Something like this. Right? So you will have to remember these formulas. Actually, the formula we were taught there was this. Okay? Plus 3AB and A plus B. So, both of them used to come plus. And in minus, it used to come like this. Minus 3AB, A minus B. So, there is a negative and a positive. Okay? Rest, you can remember as you remember. Now, here the formula will be A cube plus B cube plus 3A. square B plus 3AB square. So far, we have NCERT and extra questions. The previous questions were not in NCERT. So even if the power increases, don't worry about 500, 600, 1000. Now, here it is. 1 upon 27. Here, 27 iota cube. Now, see, it's 2, 3. Let's do one more step. No problem. Okay? The kids should understand. It's 3 to 3. What's the answer here? 9 iota square 1 by 27 now remember what is the value of iota cube minus iota sir we can write directly yes we can write why not then here it came this cancelled only iota was left and this came our minus 9 so real part with real this came real part with real and this came imaginary part minus 26 iota see, minus 27 iota plus iota minus 26 iota you solve this LCM, you will get the answer tell me the answer in the comment you will take LCM, LCM is only for real part cube also came, what will we do if power 4 comes you also know that, see if not, see 1 minus iota's power was 4 so square it and whole power 2 here power 5 will come So 1 minus iota square into 1 minus iota cube. You can do this too. There are many other ways. a minus b whole square. So a square plus b square minus 2ab and square is above. Now iota square's value is minus 1. This will be deducted. What is left? Minus 2 iota whole square which is 4 iota square. And 4 iota square's value is minus 1. So the answer is minus 4. So therefore, the value of z is minus 4 plus 0 iota. Is it clear? It's correct. Now see, it's been half an hour since we have read this chapter. And what we are doing is standard form. Because the main game of this chapter is standard form. Come on. The next question comes. See this. Now you have to convert this also in standard form. On top, identity is being made. a plus b and a minus b. Make it. So this is a square and b square. It's the same question of MCRT. In upon. I'll open bracket below. So, here comes root 3 plus under root 2 iota minus root 3 plus under root 2 iota. May God bless you. Root 3 is cut from root 3. Up comes 9 minus 5 iota square. Below comes 2 under root 2 iota. Up comes 14. How does 14 come? What is the value of iota square? Minus 1. Plus is 14. Upon is 2 under root 2. Iota. 7 upon is root 2 iota. Neither can I leave root 2 below, nor can I leave iota. Even if I leave root 2, I can't leave iota. So what will we do? Rationalize. Good. So our answer is 7 root 2 iota upon 2 iota square. And what is the value of iota square? Minus 1. Therefore, my z value is minus 7 under root 2 iota upon 2. This is my answer. I am sitting here with hope that you are doing it yourself. What do you think? Watch the next one too. Don't watch the next one. The next one is yours. Okay? It is a little weird to watch now. But very easy. Now watch. For any two complex numbers, whenever any two complex numbers come, Z1, Z2, Z3, then consider Z1, Z2. You cannot consider them in numbers, they are constant. There are many questions ahead. This is not a question. We will go ahead and there will be many questions today. So look, first of all you have let Z1. Let's say we let Z1, A plus Iota B. And we let Z2, let's say X plus Iota Y. Right? A plus Iota B. and x plus iota y. Now, to prove, we have to prove re of z1 into z2. Now, what does re mean? re means real part of z1 into z2. That means, multiply z1 and z2. Write down the real part of it. is equal to, what is it saying? real part of z1 into real part of z2 Minus, IM means imaginary. IM means imaginary part of Z1 and imaginary part of Z2. They say it should be equal. They say it should be equal. We have to prove this. It will be proved. It is not a matter of tension. Keep watching. Keep watching. First of all, let's pick up LHS. Okay? So, we have picked up LHS, which is real part of Z1, Z2. Now, we have to multiply Z1 and Z2. We have to multiply Z1.Z2, which is A plus Iota B into X plus Iota Y. We have to multiply this. Multiply this. What is this? AX plus AX. a y iota plus b x iota plus b y iota square multiply iota square value is minus 1 so this is a x a y iota b x iota and here comes minus b y so real part with real plus imaginary part with imaginary Standard form will be like this. This is the standard form. After multiplying, what do we have to convert first? Standard form. Whatever happens, it is standard form. Whatever happens, it is standard form. Okay? So, Z1 into Z2 is here. But, now what do we have to find? Its real part. So, real part of Z1 into Z2. We just have to write real part. What is that? Ax minus By. I wrote this. I just had to write real part. So, I wrote real part. Now let's talk about RHS RHS says that you have to write Real part of Z1 Real part of Z2 minus imaginary part of Z1 and imaginary part of Z2 Real part of Z1 See what is the real part of Z1 A and what is the real part of Z2 X minus imaginary part of Z1 B and imaginary part of Z2 is Y this is our answer check if both are equal LHS is equal to RHS this question is done and your question is about RE and IEM to get knowledge right? last year's one shot was 37 minutes 37 minutes check it 37 minutes as far as I remember 37 or 57 it was 37 so in that time I had increased the complex number here it is all detailed because extra questions are coming in between till now NCRT finishes like this I finished NCRT in two lectures in paid batch but after that I did many lectures on extra questions and from there I have brought maximum questions so paid batch also has done all the questions so let's go then this question is also of miscellaneous we are going concept wise if you look carefully at our NCRT first exercise is not completed there is a topic of multiplicative inverse and this is the first question of miscellaneous because this is also evaluate evaluate means nothing, standard form what does it mean? convert to standard form let's do it, no weird question iota's power is 18 straight square's power is 9. 1's power is 25. Iota's power is 25. Iota into Iota's power is 24. Why did it come up? Minus 1's power is 9. Iota's power is 24. Iota's power is 12. This is minus 1. Minus 1's power is 12. Answer is 1 by Iota. cube. Which I can write as 1 by iota minus 1 cube. And you know how to write its whole cube. I just gave you the identity. What was it? Do it yourself. A cube minus B cube minus 3 A square B plus 3 A B square. This is it. So 1 by iota cube, now you remember, minus 1 by iota. Meaning, meaning, meaning, let's go. Meaning, what is the value of iota cube, son? Minus iota, you remember? Minus 1, 1 by iota square is, and this is 3 by iota. So 1 by, let's go, let's go. Minus 1 by iota, plus 3 by iota. This is minus 1, minus 3, and this is also minus 1. Because the value of iota squared is minus 1. See, your calculation is your own. Okay? Now, everyone doesn't do calculation in the same way. I have to do it very basically. Because, obviously. Every child is not equal. So, this is 2. But, this is 2 by iota. So, it is not a standard form. 2 by iota is just rationalization. Don't do it for everyone. Okay? So, what we have to do here is rationalization. So, multiply by iota and divide by iota. So, this is 2 iota upon iota square plus 2. 2 minus 2 iota is my answer. Right? 2 iota and this is minus 1. Brother, according to me, 14-15 questions have been asked. What are we doing there? Standard form, standard form, standard form. There is one more question, son. Standard form. It is from NCRT. What do you say? Should it be given in homework? Should it be given in homework? So this question is your homework. Here, first take it with LCM and multiply it. And if the iota comes down, then rationalization. This is one question in the homework. It is a question from NCRT. And you have to tell me the answer in the comment. Please. Okay? The correct answer will be pinned. So, this is our standard form game. Okay? Now, the topic is multiplicative inverse. I had read two words in my childhood. One was multiplicative inverse and one was additive inverse. Look, additive inverse means to change the sign. 5's additive inverse is minus 5. And minus 6's additive inverse is 6. Multiplicative inverse means reciprocal So we are told that multiplicative inverse of 2 minus 3 iota. So the answer is 1 upon 2 minus 3 iota. It is that simple. 5's multiplicative inverse is 1 by 5. 6's multiplicative inverse is 1 by 6. So 2 minus 3 iota's multiplicative inverse is 1 upon 2 minus 3 iota. But the answer is there. But it is not in the standard form. Whether the question is said or not. You have to convert your answer into the standard form. So if we look carefully here, then what we have to do is rationalization. What we will do here is rationalization. So here it comes, 2 plus 3 iota divided by a square minus b square. So 2 plus 3 iota divided by 4 minus 9 iota square. Right? Let's go. So here it comes 2 plus 3 iota whole divided by 13 converted into standard form 2 by 13 plus 3 by 13 iota You are doing this whole process from the last 15-16 questions But understand this, multiplicative inverse means reciprocal To do it in standard form, you are doing rationalization Next question is multiplicative inverse of this. So, this one should be given in homework. You will tell in the comment. Quickly, dear children, quickly answer this in the comment. How can we do multiplicative inverse? Right, children? Now, listen to me carefully. One of your studies is complete. That is standard form conversion. But, it is not easy yet. Why is it not easy? Because whatever you will study in life, means whatever you will study in this chapter's life, We will study Argon Plane, Modulus, Conjugate. Oh, great concept. Conjugate. So, for all of us, we need one thing, which is Standard Form. So, until the question is not in Standard Form, we will not be able to proceed. Maximum questions. So, when we are moving forward, we will study Standard Form. So, what is your work now? This is your high. But now, you can sit here and complete 4.1 from the point of view of the new NCRT. If you have old NCRT then 5.1 Complete 4.1 with examples It will be completed with love The questions that you have left for 2-4 months You can do it easily Complete all 4.1 with examples Then we will come to miscellaneous But before going to miscellaneous We will see some conclusions and formulas Come let's see Now we will talk about Argand Plane Now Argand Plane Before that you have to revise what was Cartesian plane In Cartesian plane there was one axis and one axis Which we called x axis and y axis Both were real axis there were 4 quadrants and it used to plot the real quadrants Now what is an Argand Plane? Here there are only two axes but the one axis will be imaginary Meaning? Like Cartesian Plane Let me make one for you Like this and this So, this is x, this is x dash, this is y and this is y dash. We call this axis real axis and this axis imaginary axis. Now, there is a complex number. Let's say. 2 plus 3 iota A complex number is Let's say 2 plus 3 iota And here is 2 And here is 3 So this is the plot This will be 2 plus 3 iota This is the point Now here the coordinates Here in the Argan plane The coordinates will be Like 2 comma 3 So this will be real Obviously And this will be imaginary 2 comma 3 means Here is 2 plus 3 iota If it is written for example, 4 or minus 4 plus 5 iota. So, let's assume this is minus 4 and this is beta 5. Okay. Let's assume. So, this plotting will be here. We will call it minus 4 plus 5 iota. This is what z is representing. Okay. Assume it is written here minus 4 and minus 3 iota. So, this point will be minus 4 minus 3 iota. This is beta z. So, it means. The story is simple. One second. The story is simple. This plane is called R-Gantt plane. Now, one axis is real and one is imaginary. Whatever is plotting is of a complex number. Here, introduction of real number is not there. Obviously, if you just want to plot 3. 3 plus 0 iota. Then it is plotted here. 3 plus 0 iota. If you want to plot, for example, you won't get plotting, this is a topic which is mentioned in your NCRT, there is no question on it, but it is a very important topic to go ahead. This year it may not be used, it may not be used in the 12th, but it is used in studies. So here comes Y, if someone asks you to plot, let's say 4 iota, then 4 iota is only this, only on the imaginary axis. So this is our Argan Plane. Now, why did I teach you this? Its use will be a little bit in the modulus of a complex number. First, read modulus of real number. What was modulus of real number? If we, for example, absolute value, the mode of minus 3 would be 3, and the answer of 3 would be 3. So it absorbs minus. Now when it absorbs minus, we call it real number's modulus, i.e. absolute value. Now, let's understand what is the modulus of a complex number The definition of the modulus of a complex number is Distance of a complex number from origin in an Argand plane Look at the Argand plane When you plot any complex number on an argon plane, the distance from its origin is called its modulus. If z is a plus iota b, then the modulus of z will be under root a square plus b square. You have to remember this. Now, where did this come from? If I explain it to you with 2 minutes extra, then it is a very simple method which is kept in the definition. For example, let's say this is Z. Z means A plus iota B. So its coordinate is A comma B. And this is our origin. Now the distance from here to here, I will call it modulus of Z. That means if you apply the distance formula, if you apply the distance formula, then mod Z will be, Let's apply your distance formula here. a minus 0 or 0 minus a. It's the same thing. Its square. And b minus 0 is its square. So the answer is a2 plus b2. I have derived this for you. Otherwise, you have to remember it. For example, if I have given a z, minus 3 plus 4 iota, then you don't have to do anything. The z will become modulus. Its square plus its square. So the answer is 25. That means 5 units. You have to do it for iota's neighbor, not the whole. That means 4 square. Don't do 4 iota's square. Kids make mistakes. square it by 4 iota and write minus here. That will be wrong. So, how did it come? The distance from the origin of a complex number is called modulus. Is it clear? This is the reason. This is the definition. This is the formula. This is the example. Got it? Now we are going to do a lot of questions on this. Now the use of argon plane was here. If I didn't understand this, how would I have explained it? Then comes conjugate. Now conjugate of a complex number is nothing. Only and only mirror image. I had read a topic in 9th standard. I will revise it. What is this? Mirror image of a complex number. Now we will prove it and show you. Let's assume that if z is x plus iota y then the conjugate of z will be x minus iota y. For example, if Z is given somewhere, then Z will be conjugated as. This means that only change the sign of the imaginary part. Change the sign of the imaginary part. If there is a plus sign, then there is a minus sign. If there is a minus sign, then there is a plus sign. Why is this happening? Now I will take you to the argon plane and explain the definition. Assume this is an argon plane. What will mirror image mean? Assume this is 3,4. Mirror image of complex number treating x axis as plane mirror. This was a topic in 9th K coordinate geometry. I will teach the children. But if you have forgotten, you will remember a topic of science If this is a plane mirror and the object is here then the image will be made at the same distance and of the same size So if we want to get a mirror image of 3,4 If I consider this as a plane mirror If this is a plane mirror, then where will the image be? The image will be here 3, minus 4 will come Where will the image come? 3, minus 4 will come here 3,4 and this is 3, minus 4 understood? now, sign will change 3,4's complex number was 3 plus 4 iota and this is 3 minus 4 iota if suppose there was a point here minus 5 minus 6 so its image will be here minus 5 plus 6 will be here so the answer will be minus 5 plus 6 iota what was the question? minus 5 minus 6 iota its conjugate is this and this is its conjugate so if you understand this, then it is a very good thing you will remember and become superior if someone asks what is conjugate then you will not have to say that read imaginary and change the sign that is how conjugate comes out that is a trick but the real meaning of conjugate is this the real meaning of modulus is distance from origin so I have told you the real meaning how to use it if z comes, its modulus then just under root Real part square plus imaginary part square. If someone asks to find conjugate, then what to do? Simple thing to do. Just reverse the sign of imaginary part. Now I am telling you a small formula. I will prove it further. Z, the square of mod Z, will be equal to Z into Z conjugate. The square of mod Z, will be equal to z into z conjugate. This is a very good formula. It will be used a lot. It will be used a lot. Is it clear? Let's move on to the questions. Now we have to find the conjugate. We will find the conjugate. But first of all, we will do the standard form. We can write the direct conjugate but our answer will not be valid. Either you can take conjugate first and simplify it or you can simplify it first and then conjugate it. I always prefer simplification first and then conjugate. So first we have to convert it into standard form. So this is 3 minus 2 iota and this is 2 plus iota. Multiply it. What is it? 6. This is 6. plus 9 iota minus 2 iota minus 6 iota square whole divided by here 2 goes there minus iota plus 4 iota minus 2 iota square now do it in standard form this is plus 6 this is 12 plus 7 iota divided by this is 4 plus 3 iota what we have to do in standard form is rationalization we did rationalize it 48 minus 36 iota plus 28 iota minus 21 iota square a square minus b square 48 plus 21 iota square minus 1 48 plus 21 69 minus 8 iota How much is there in the upon? 25. Now see our standard form. Let's see here. What is our Z? What is the standard form? 69 upon 25 minus 8 upon 25 iota. What we have to get is Z's conjugate. Conjugate means Z bar. Nothing to do. You have come to the standard form. Now you have to change the sign of the imaginary part. This is our answer. This is the end. I told you that 98% content is the standard form. So we are doing only one thing from here to here. That is the standard form. So we reached here. After this, what did we remove? Its conjugate. It will say conjugate, conjugate modulus. Modulus or ask something else. Using conjugate and modulus, we will do that too. See more. Now as I am saying, what to remove? Its modulus. So modulus is an obvious thing. It cannot be removed without the standard form. conjugate will still be there but we have to solve it in standard form means if I solve all these by 3 plus 2 iota 2 minus 3 iota 1 minus 2 iota 2 plus iota then I get this answer first conjugate then solve but here I can't do this that I do modulus above and then down here we have to do it in standard form so for now my z is 1 plus 2 iota and 1 minus 3 iota To do it in standard form, I have to do rationalization. So Z is 1 plus 3 iota plus 2 iota plus 6 iota square upon 1 minus 9 iota square. This is 1 minus 6 iota plus 5 iota upon 10 iota. To do it in standard form, I have to do minus 5 by 10 iota plus 5 by 10 iota. So my Z value is minus 1 by 2. plus 1 by 2 iota but this is z we don't want z, what do we want? z's modulus and how will it come? under root of real part square which is 1 by 4 and imaginary part square is also 1 by 4 so 1 by 4, 1 by 4 is 1 by 2 and what happens after that? under root 2 is the answer ok? sir, is this only a standard form? No, there is a polar form also, I just remembered, I remembered because it has been deleted for many years. There was a polar form also, it was removed only when the question was asked. By default, there was a standard form, there was a polar form also, whose formula was R cos theta plus iota sin theta. Okay? Trigonometry was used there, but now the trigonometry has been removed from the whole complex number. That means, when it was deleted during Covid, it has been removed from that, it has not come since then. Let's go. Now what we have to do here again? Modulus. One is mine and one is yours. To get modulus, first of all we have to do standard form. I am now practicing modulus for you. Big questions are also coming. So here comes 1 plus iota whole square. Below comes a square minus b square. a square plus b square plus 2ab divided by 2. iota square will be 1. 2 iota divided by 2. Z value is only iota. You can say that the value of z is 0 plus iota. Standard form. Now, what will we get? Modulus. That means 0 square 0, 1 square 1. So, the answer is under root 1. Right? Done, done, done. And see, what is he saying? The next part of this was this. You must comment on this answer. What is the modulus? Say 6. Make it sure. Quickly. What is he saying? The real number of x and y. We have to find the value of x and y if this is is the conjugate of this. Meaning, x minus iota y, 3 plus 5 iota, is equal to minus 6 minus 24 iota, conjugate. Right? We will solve. This is 3x, this is 5x iota, minus 3y iota, and minus 5y iota square. is equal to minus 6 plus 24 iota. Standard form. This is 3x plus 5y. Real part with real. 5x minus 3y iota. Imaginary with imaginary. And this is the first concept. When two complex numbers are equal, then real part is equal to real. And imaginary part is equal to imaginary. Therefore, 3x plus 5y will be equal to minus 6k. And 5x minus 3y will be equal to 24k. Elimination substitution. Take out the value of x and y. Answer the comment. If you are a good kid, you can do it. No problem. Is it correct? Is it correct? See more. Now, as the question came here. I will make the question a little visible. Z1 is given as 2 minus i. Z2 is given as 1 plus i. You have to find modulus of Z1 plus Z2 plus 1. Z1 minus Z2 plus 1. But first, the modulus cannot be found without the standard form. So first, put the value of Z1. What is it? 2 plus 2 minus i. What is the value of z2? 1 plus iota. Along with that, it is plus 1. The value of z1 is 2 minus iota. Minus z2. The sign will change. And here it is plus 1. Now see, cut the sign that is being cut. iota is cut from iota. And here 1 is cut from 1. So, our 4 came up. Below came 2 minus 2 iota. You can also take 2 common and cut it. Here it is 1 minus iota. But we will first convert its standard form and then do the modulus. So come and see. 2, 1 minus iota, multiply by 1 plus iota, divide by 1 plus iota. This is 2, 1 plus iota, divide by a square minus b square. Right? Up comes 2, 1 plus iota, and how much is 2 below? Because the value of iota square is minus 1. This will be cut. Modulus 1 plus iota. Now what is its modulus? Square of real part and square of imaginary part. So the answer is under root 2 units. Right, right, right. It is simple. Look at it more. Here it is given Z1, Z2. We have to get its value. Here it will come and here it will come. We will correct the question. It is a question of NCRT. So, like I am doing the first part, the real part of this. You have to find the real part of what is z1? 2 minus iota. Okay, here we are using conjugate, not modulus. 2 minus iota. What is z2? Minus 2 plus iota. And divide by z1's conjugate. If I conjugate, then its sign will change. What did you write for 2 minus iota? 2 plus iota. You must have remembered conjugate. Change the sign of imaginary part. Now, solve this. Real means real. I have already told you this. So, minus 4 plus 2 iota plus 2 iota minus iota square divided by plus 2 iota. This is what we got. Okay? I won't need this anymore. How much did we get? Minus 3 plus 4 iota upon 2 plus iota. Convert this in the standard form. And write the answer of the real part at the end. What is it? Minus 6 plus 3 iota. plus 8 iota minus 4 iota square divided by 4 minus iota square Real part, this is minus 2 plus 11 iota below this is 5 and below this is 5 Now tell me the answer to the real part, minus 2 by 5 If you ask imaginary, then 11 by 5 See, I am saying the standard form, till now I have done 28 slides Leave 2 slides, 3 slides, 4 Form the standard in 25 questions 24-25 questions and you are doing standard form in all of them. How much more practice will you do? So, you did standard form in 24-25 questions. In the end, if you said real, then real. You said modulus, modulus. You said conjugate, conjugate. Imaginary So here is imaginary Here is bar, you will comment the answer I am saying this again, I am giving you this so that you can practice yourself and not get spoon-feeded And all the questions I am giving you are from NCRT That is why I am giving you this, because they are very basic questions As you are saying to find the modulus of this, to find the modulus of this you will have to use standard form In standard form you can do different rationalizations or you can also crack its LCM here comes 1 plus iota whole square and 1 minus iota whole square LCM is here a square plus b square plus 2ab this is a very basic question and I am going fast now we will come to a very exciting question you will enjoy doing that more iota square divided by 1 is 2 iota iota square divided by 1 is minus minus plus 2 iota and how much is 2 below So, the answer is 4 iota and 2 below. So, the answer is 2 iota. This is your z. Now, what we have to find out is the modulus, mod z, the square of imaginary part and the square of real part. So, the answer is 2 units. Okay? Understood? I have presented you all the questions of different varieties. Check it out. Now, we have to see these four questions. In these four questions, can you see? 1, 2, 3, 4. There is a special thing in all these four questions. Look at it carefully. I will give you a complete track. Look at this question, look at this question, look at this question, look at this question, look at this question, look at this question. There are five questions. There is only one special thing in all these five questions. Can you guess what it is? I will show you. You have given a plus iota b. The question is a square plus b square. Here you have given a plus iota b. The question is a square plus b square. c plus iota d. The question is c square plus g square. e plus iota f. The question is e square plus f square. g plus iota h. g square plus h square. a plus iota b. a square plus b square. x minus iota y. Correct the question. It will come here also. x square plus y square. Here x plus iota y, what did you ask? x square plus y square. Whenever you are told that x plus minus iota y is given in the question and in the answer you have to get this proof or value of this. It should be plus here. Then we will always follow the same concept which I just told you. Okay? Let's go. What is the concept? Get the conjugate. So we will start from here. See, all the questions are the same. You wrote this question. All the equations are same. There are 2-4 NCRT and 2-4 BHAAR. This is the first equation. What to write? Taking conjugate both sides. Write this and take conjugate. Conjugate means real part's sign change. Sorry, imaginary part's sign change. Up came A minus Iota B, down came A plus Iota. And then what do we do? We multiply it. Multiply the first and second. Come, multiply the first and second. This is x plus iota y. This much common sense will be there. Let's direct it. a plus b, a minus b. What is it? a square minus b square. We will talk about this. Up comes a square minus b square. If you talk about the bottom, then also by chance what is coming? a square minus b square. What is the value of iota square? Minus 1 What is x square plus y square? Whenever this happens, we have to do this only. In the 5th question, we are going to talk about only one thing. This is completely cut. What is the answer? 1 Without wasting time, let's move on. Yes, x minus iota y. What do we need? The whole square of x square plus y square. First write this question. This is a minus iota b and c minus iota d. Now see, it is reflecting here also. What do we want if it is a minus iota b? a square by b square. What do we want if it is c minus iota d? c square by d square. We have to do this. What is it? Taking conjugate both sides. Conjugate both sides. What is it? x plus iota y. This is a plus iota b. And this is c plus iota d. What do we do with both? multiply what is this? minus b square here also it comes a square minus b square here also it comes a square minus b square what is the value of iota square? minus 1 here comes x square plus y square and this is our a square plus b square upon c square plus d square answer comes it asks for proof but it says here is square So don't do anything, you can do the last step from your side also. Squaring both sides. This is our answer. Next one is also same as this. And this is such an important question, it comes in the paper. See this is the question. Its plus is little bit higher, rest is fine. A plus iota b, what do we need? a square plus b square. Look carefully, what do we have to do? A plus iota b, we wrote the question first. x plus iota's whole square. 2x square plus 1. This is the first. What will we do kids? Taking conjugate both sides. Conjugate both sides. Here comes a minus iota b. x minus iota whole square. Understand. x minus iota whole square. But below there is no iota. Below it is all real. How will its sign change? The conjugate of 3 is 3. Because it is real. And below it is real. Children make mistakes and write minus in flow. So here comes our second. What will we do? Multiply. Multiply and the answer will come. a square. Do it like this. x plus i. x minus i. Up comes square. Down comes 2x square plus 1 square. Solve it and the answer will come. Prove it. Look here. Will you do homework? This is the first equation. Conjugate it. Multiply it. This will be proved. I will do this. So far, all four NCRDs have said that this is our outside question, this is our exemplar question, and I cannot tell you how many times this has been mentioned in the paper. Because I myself don't know. I have seen this question so many times in the paper. Because there is one more extra thing in it, and that is this. This is the interesting thing. Okay, son. Now you have to look at it a little carefully. The question is amazing. Okay, now see. What is it? The first thing is the same. A plus iota B. C plus iota, C minus iota. This is the first. Taking conjugate. We have to do that. A minus iota B. C minus iota. C plus iota. This is the second. We will multiply. We have to do that. What is this? A square minus B square. C square minus iota square. C square minus iota square. If it is cut, then the value of A square plus B square is 1. This is the first part. In the second part, you need B by A. It's division. So what we will do is divide these two. We did multiply earlier. Here we will have to divide. Divide first and second. Very important game. You have to divide first and second. A plus iota B. C plus iota C minus iota. Whole divide by. Whole doesn't mean like this. It means here is different. Here is different. This is different. a minus iota b here it is c minus iota upon c plus iota right? both of them divided now see first of all this will multiply this will multiply a plus iota b and a minus iota d has anyone from you how many children have heard the name componendo and dividendo comment raise your hand how many children eh? I have heard the name of Componendo and Dividendo When I was in 10th standard If you have studied 10th standard last time I have said Componendo and Dividendo so many times in live lecture Means I don't remember it myself And I had told something there So here you have to apply Componendo and Dividendo There is no other option In the whole 11th standard There are 10 to 12 questions Where the question will not be solved without Componendo and Dividendo So we will solve this question with Componendo and Dividendo Now what is C and D? I will write it down so that you remember when you see the PPT. Compo nendo and dividendo. What does this mean? If A by B is equal to C by D, two ratios are equal to each other, then if I apply C and D, then numerator plus denominator, numerator minus denominator. numerator plus denominator numerator minus denominator Up will be numerator plus denominator and down will be numerator minus denominator Look here numerator plus denominator and numerator minus denominator here comes c plus iota square plus c minus iota square c plus iota square minus c minus iota now above is cut from this and below is cut from this above is left 2a and below is 2b iota If you see above, c square plus iota square plus 2c iota. c square plus iota square minus 2c iota. Below, c square plus iota square plus 2c iota. Minus c square minus iota square plus 2c iota. Opened the bracket. Now cut whatever is being cut. See, first 2 is cut from 2. If you look here, then 2C iota to 2C iota, C square iota square, C square iota square. I have a little space left. A upon B iota. If you look here, then twice of C square plus 1 is here. Okay? And if you look below, then we have 4C iota. This is also cut. Reverse it. B iota upon A. And here we have 2C iota upon C square plus 1. Iota from iota is also cut. This is what we had to prove. Check it out. Now the question is why it is important. You must have understood it. Because here this is the important game. We understood this because we had to do it like the last four questions. Whenever there is a divide, divide both. And most probably the component dividend will be there. And let me tell you one thing. This is a one of a kind question. You won't get this question by searching. This is the only question. You won't get this question by searching. This is such an important question of your exemplar. And I can't tell you how many times I have written it in the paper. Trust me, prepare it very well. I am a competent dividend. Go back and rewind it a little bit and understand what I just explained. Right? Let's move forward. So this topic is clear for you. Next question. Now see, you will think this question is like those four or five questions. But this question is not like that. See, why? Because here is x plus iota y. But here is not x square plus y square. And this question is the most important question of our NCRT. Very important question. You are my old friends. This is a very important question. This question is also very important. Rest of the questions are very important. Why? Because they will come after 1 month. All the questions were good. But we didn't get any bad questions. To make a base for all NCRT, we have to do some basic questions. But since this question has started, this question is very important. The previous 4 questions are also very important. And this one has come in such a small number of papers. Or if you have this chapter in your UTPT, for example. Then comment that this question has come. Many have come and many more questions are going to come. So here comes whole cube of a plus, that is, x plus iota y. U plus iota v and we have to prove this. So here it will take the conjugate process. Here the same normal process. And what is the normal process? Standard form. So see, we will do the standard form of this. Here comes a cube plus. b cube plus 3a square b plus 3ab square and here u plus iota v. Okay. x cube is x cube. iota cube is minus iota plus 3x square y iota and this is minus 3xy square because iota square has minus 1 value. And you pay attention. I am also doing the standard form here. So, this is the real part. And here, by taking iota common, this is our imaginary part. Okay? For those who did not understand this step, I will tell you again. See, this is real in one bracket. I took iota common from both of them. And imaginary came in one bracket. Real plus iota imaginary. Okay? Now, if both these complex numbers are equal, see, how much is the first point to remember coming in? If both these complex numbers are equal, then the real part will be equal to U and the imaginary part will be equal to X. Take common from this. What is X square minus 3Y square equals to U? So X square minus 3Y square equals to U by X. Why take common from below? Because I am demanded in the question. We need u by x. Here what is common? y. Here 3x square minus y square equals to v. So here 3x square minus y square is the value of v by y. We will check. Did it come or not? Now see, we need the value of u by x plus v by y. If this is first and this is second, then what will we do with both of them? Add first and second. And our work should be done. So u by x plus v by y u by x plus v by y we will add x square 3x square. 4x square. Minus 3y2 minus y2 minus 4y2 Take 4 common, kids. This is our answer. Excellent question. Chummeshwari question. The more you put your heart into it, the less it is. The less it is. Shaloo. Next question. Speechless. Such a question. Such a question. Mind-blowing question. I mean, I don't know the words. I praise it so much. This means that Shah Rukh Khan is a complex number. If Shah Rukh Khan is a complex number, then he is a Koshan. I take my words back. Shah Rukh Khan is not there. He is Diljit Dosanjh. He is everywhere. Take the movie South, there is Diljit Dosanjh. If it is being shown in a foreign country, there is Diljit Dosanjh. If a girl's crew movie is being shown, there is Diljit Dosanjh. If there is an actor... He is doing some promotion and Diljit Dostandhya is coming there. So here, Diljit Dostandhya of every paper is this question. This question was 6 marks in my exam. When I was in 11th standard. I am telling about finals. And after that, I have seen at least 600 times in papers. I must have seen a lot. So, brother, hold your heart. The question is very important. Here, the concept will be applied. I will prove the proof that I told you. Okay? First, here, conjugates are above alpha. Okay? So, it is the question of your NCRT. Alpha and beta are two different complex numbers. The value of mod beta is 1. You have to find the value of this. Modulus of beta minus alpha upon 1 minus alpha conjugate beta. We had read that the square of mod z is equal to z into z conjugate. I will prove this and remove it. You can take a screenshot. a plus iota is also z. Modulus will be a minus... a square plus b square so the square of modulus is a square plus b square and z into z conjugate means a plus iota b and conjugate is a minus iota b so what is a square minus b square? a square plus b square so these two are equal so the square of z mode is equal to z into z conjugate take a screenshot and I will remove it so this is the equation for this equation Okay? Okay? Or if I see, this is a PDF, right? Yes, it will be possible. It will be possible. We will make it small and put it on the side. You see, I have made this place in your notes. Okay? Zoom in and understand. This is proved. Now, we will see here that mod z square is equal to z into z conjugate. So, mod beta minus alpha 1 minus alpha conjugate beta square Z is the same. If beta alpha is a complex number, then this will also be a complex number. So, beta minus alpha, 1 minus alpha conjugate beta Z and into means Z conjugate. Now, focus on RHS. LHS will continue. Beta minus alpha. 1 minus alpha conjugate beta. Conjugate all of them. Beta conjugate, alpha conjugate, 1. Conjugate of conjugate. That will come. And this is here. You understood this, right? That I will conjugate again of any conjugate. So that will come. A plus iota b conjugate. A minus iota b. If I conjugate again, then a plus iota b will be found again. This is here. So see. we will multiply it, till now we have not done anything we are doing only the basic things here it is beta into beta conjugate alpha conjugate beta alpha beta conjugate alpha alpha conjugate here it is 1 minus alpha beta conjugate alpha conjugate beta, look carefully, plus alpha alpha conjugate beta beta conjugate, LHS is the same now when z into z conjugate Modulus z square. So beta into beta conjugate. Modulus beta square. this formula is applied again alpha conjugate beta alpha beta conjugate this formula is applied here here it is the same here it is the same here it is the same what was happening in LHS beta minus alpha 1 minus alpha conjugate beta square that is the same mode beta and the value of mode beta is 1 so this is our 1 minus alpha conjugate beta alpha beta conjugate plus mod alpha square same below 1 minus alpha conjugate beta minus alpha beta conjugate mod alpha square and mod beta square then what happens? 1 normal means the whole question is being done with the same formula now now cut them out, see the same is there so beta beta minus alpha 1 minus alpha conjugate beta square is 1 square will be under root and under root value is 1 why not plus minus? because modulus answer can never be negative because it is distance so answer is 1 very simple I will revise it again if you want it deeply then rewind it remember this formula any complex number square is equal to its conjugate I have conjugated everything. Alpha conjugate of conjugate, alpha again. I have multiplied the brackets. Now, Z into Z conjugate, mod Z square. Z into Z conjugate, mod Z square, here also. Now, what is the value of mod beta? 1. I put it everywhere, cut it, and the answer is 1. So, there is nothing special in the question. Now, you will say, sir, at your time, this was of 6 number. Yes, why? Because the pattern was 1, 4, 6. Sometimes, it was of 4, and if the paper was easy, it was of 6. So, my 11 paper was easy. Can say that. Ok. Let's go. It was a good question. Now see, this is our last question of NCRT. I have given you many extra questions. Now there will be more extra questions. Don't worry. Now this question came. 1 plus iota, what is the power of 1 minus iota? 1 minus iota. What did I say? Power m is equal to 1. You have to find the least, i.e. the smallest, positive integral, i.e. integer value of m. You have to tell the smallest positive integer value of m. We can't do anything without channel form. Come on. Here it is. 1 minus iota is 1 plus iota. Power is m is equal to 1. So here it is. a plus b whole square. a square minus b square. Whole power is m. Here it is 1. a square plus b square plus 2ab. Divide by 2. Whole power is m is equal to 1. iota square will be 1 this will be 1 iota power m is equal to 1 so m is equal to 4 now understand why 4 is here if i put iota power 1 see it should not be positive, it should be greater than 0 iota power 1 is iota iota power 2 is minus 1 iota power 3 is minus iota so 4 is 1 how? iota square is square So, the answer is 1. So, there is no value smaller than 4. There is a big uncountable value than 4. And positive, although there will be a lot of values in negative, but what we want is positive. So, the answer is 4. Okay? Clear? Now, we will move on to its other best extra questions. Like this question. You will not get this question in NCRT. You will not get this question in NCRT. You will get this question in exemplar. See it properly. What is it? Where does Z lie? Where did Z lie? Oh, sorry. Where does Z lie? Now, lie means where does it lie on your argand plane? First quadrant, second quadrant, third quadrant, fourth quadrant, x-axis or imaginary axis. We have to know this. Okay? So whenever there is such a question, now see, I'll show you the line. This is Z. In the next question, we can see Z. In this question, we can see Z. I'll give this homework. In this question, we can see Z in MCQ. This is MCQ in the eyes. This question also doesn't have z. See, I will show you the z question. Here is z. Conjugate has come. Here is z. And this is the same question. Okay? So, all these z questions, all these z questions, you have to do one thing. Let z be a plus iota b or x plus iota y. Do anything. So, we have let here. Let z is equal to x plus iota y. Now I am letting you play x plus iota y because we are playing a game of quadrants. So it is easy for you to understand x, x and y. By the way, if you do a plus iota b, it is not wrong. Now here comes the modulus. x plus iota y minus 5 iota. And x plus iota y plus 5 iota is equal to 1. Right? Now see. We will do it in standard form. Here comes x plus y minus 5 iota. x plus y plus 5 iota. And is equal to 1. Now, let me tell you something. Whenever it is written like this, let's assume z1 upon z2 is equal to 1, then the conjugate of z1 is the modulus of z2. It can be equal to this. Okay? So, this is possible. Look again. This is possible. Take a screenshot. Come on. Now, what are we doing here? x plus y minus 5 iota and here x plus y plus 5iota modulus Modulus is the square of real part and the square of imaginary part and here also square of real part plus square of imaginary part square root is squaring both sides so here comes x square y square plus 25 minus 10y x square y square plus 25 plus 10y see this x square y square 25 x square y square 25 is cut minus 10y is equal to 10y here it comes minus 20y is equal to 0 so y is 0 if y is 0 then my z will be x plus 0 iota and remember where did I plot 3 only x will be there, it is not imaginary so it will plot on real axis so z lies Z lie on X axis or you can write on real axis. Z will lie on real axis. If it comes to 0 plus 5 iota, understand, it got dark, I will come back. If X comes to 0 plus 5 iota, then it lies on imaginary axis. If value of Z comes to 3 plus 4 iota, then it lies on first quadrant. 3 minus 4 iota. I would have lied in the second. Basang came. So, it is a game of quadrant, Babu brother. Right? So, here it happens. You must know. Here it is plus plus. Here it is minus plus. So, I may have told you wrong. So, 3 minus 4 iota. So, he will lie here. So, here it is minus plus. Here it is minus minus. And here it is plus minus. Here it happens if iota is not there. And here it will happen if real is not there. Okay? Let's see the next question. Show that the equation does represent a straight line. Represent a circle. Locus of a straight line. This word is locus. Locus means equation. Group of points. Okay? Now, again, Z has come here. Don't worry at all. So, let Z be x plus iota y. So, x plus iota y plus 1 minus iota. x plus iota y minus 1 plus iota. Standard form. Standard form. What is it saying? X plus 1 is one. And Y minus 1 is one. Here. X minus 1 is one. And Y plus 1 is one. Same question is going on. Like the previous question. Now take modulus. How will modulus be? Square of real part. Plus square of imaginary part. Here comes square of real part. plus the imaginary part of the square squaring both sides so a square plus b square plus 2ab the root is removed I have opened the whole square of these too ok, this is it now see, cut whatever is being cut x square 1, y square 1 this is 2x plus 2x minus 2y minus 2y equals to 0. This is 4x minus 4y equals to 0. So, the equation is x minus y equals to 0. Tell me. Since this is a linear equation in two variables, which represents a straight line. Write the statement. Since this is a linear equation in two variables, what does it represent? It always represents a straight line. So, this is the equation of a line. This is the standard form of the equation of a line. Ax plus By plus C equals to 0. This was written in the 10th standard. 2 was there, pair of linear. 9 was there, linear equation in two variables. Got it? So, the flow will come automatically. Remember the z thing. What is it? z plus 1, z minus 1. You have to find z. Same, brother. z. What will z be? Let's do z again. z is equal to x plus iota y. So, x plus iota y plus 1. And here, x plus... Now, see this. These are not all one shot, just one shot. You can correct the entire NCERT. Correct the NCERT and ask extra questions. You will remember me when you go to the exemplar and say that sir has taught all of it in one lecture. Now you are watching it with your own eyes, it's your call. I always say this, the kids don't know this tagline. Salt according to taste, simplification according to status. Or according to your status, how much you can absorb in a day, you can watch the lecture. Make this lecture your friend. I am saying that it has been 19 years since I started teaching. And if I hide some questions, then the important question is coming up. Live lecture, there are a lot of things to come. In the last year, I have taken live lectures of 11th standard. I have taken live lectures of 10th, 12th, but I have taken live lectures of 11th standard too. So, there when we take a lecture of 11th standard, If you are doing marathon of 50 questions or 100 questions, then you will get complex number also. So with this your base and good questions will also get cracked. Means you will not leave any variety. Now you will have courage to attempt every difficult question. Like this question. Here it came in standard form. X plus 1, Y iota. X minus 1, Y iota. Yes, you have put modulus here. Now we will do squaring board set. We have done three questions and all three are the same. I make your work so easy. Because I find all the questions and arrange them in the same category. The work is like this. I will do it like this. I will take the squaring board set. I will find the questions here and there and first I will collect them and arrange them. Why should I arrange them? Because the questions of one order are happening in one category. Check it. Put it. Check it in full speed. This will make your work easy. I think so. x square, x square, 1, 1. y square is also gone. Very good. So here comes 2x plus 2x is equal to 0. 4x is equal to 0. Now x is 0. But y is not 0. Whatever you put, y will cut. If I put 5, y will also cut. If I put 5 crore, it will also cut. If I put root 3, it will also cut. So, the value of x is 0 and y belongs to any real number. Therefore, my z will be 0 plus y iota, where y belongs to any real number. y can be any real number. The iota's proceed is real, right? That will not be imaginary because imaginary creates real. It will be difficult. Let's go. Then comes, you will do this yourself, as we are saying that we have to prove that it is rational. Rational means to prove that it is real. The answer to this will always be real. And the answer to this will always be real. The imaginary will disappear. Now, let's do an equation here. a plus b, a minus b. a square minus b square. Multiply this. What is it? 6 plus 15 iota plus 4 iota plus 10 iota square. That's square root of 6. Minus 15 iota. And minus 4 iota. And plus 10 iota square. Now see. This is cut from this. This is cut from this. So 6 plus 6 is 12. And this is minus 20 upon 9. So minus 8 upon 9. Purely real answer. Imaginary will not come. Even in this, imaginary will not come. Okay. So just prove this. It was an easy question. Now he is saying. Now, I can give this in my homework. You can comment. You will put the value of z and tell that x is lying and y is. The question is x is equal to y or x is equal to minus y. You will put the value of z. I have done three questions like this. So, this is your homework and you have to give answer in comment section. Okay. You have done the same question. What you have to take for z? x plus iota. Then, the question is a plus iota b. c plus iota is equal to d. It is a simple thing. If A plus iota B is equal to C plus iota D, then both of them will have the same modulus. Two complex numbers are equal to each other. So, both of them will have the same modulus. This is C square plus D square. I have squared both sides. So, A square plus B square is C square plus D square. This was a small MCQ. This is the answer. I had asked the question at the beginning. The Iota's power is 39.75. That too comes in MCQ format. All of them. Now, see what is written here. See this. It is MCQ in looking. It is so interesting to see the question. They say 1 minus iota is a root of this. Okay, so the question is wrong. Equals to, with whom? With plus, right? I mean, I was looking at this. Our keyboard is equals to. Here, it should either be plus or minus. Let's take one thing and see. Something has been mistyped here. Let's take this as plus. Because equals to is with plus. Okay? So, plus should be there. But the mind is saying, it might be minus as well. I will tell you the method, then you can see the question what is it and what is not. You have to find the value of a minus b. Now here a minus b is there, so I think it can be minus b. Oh God, she will hit a shot straight. Come on, see. Here comes the whole square of 1 minus iota plus, if there is a root, then it will satisfy it, right? And here comes plus b. Write minus only because the value of a minus b will only come. So we write it minus. Okay? So, let's do minus here. It is mistyped. It is a question of exemplar. If anyone has an exemplar, then write it down in the question. So, here comes a2 plus b2 minus 2ab. Here comes a minus a iota minus b equals to 0. Right? This will cut it. And here comes minus 2 iota plus a minus b equals to 0. And this was also there. Minus a iota. So minus a iota equals to 0. So the value of a minus b is 2 iota plus a iota. This value is the same. A minus b. Is the root of this. Yes, it will be like this. There is something wrong in the question. There is something wrong in the question. Okay. Otherwise, the answer would have come by now. we are told that it is the root of this equation if there is a root then it will satisfy it that we did this, now it will be minus 2 iota we have to tell the value of a minus b that is not there here a minus b is 2, yes it is correct so a minus b plus 0 iota is there see, real part is equal to real iota is not there here And this is our 0 plus 2 plus a iota. So now the real should be equal to real. So the answer is 0. I am not sure if the question is right or wrong. If the question is this, then the answer is 0. 99.9% we are right. But it is possible that our answer is wrong because of changing the question. So rest of the work is done. See the concept. If there is a root, then it will satisfy it. And whatever conclusion comes next, it will come to the fore. Now we have to find the least value of this. We just did the question of m. So I am giving you this homework. Let me correct the question. This is the iota. And what is the power? N. We just did the question of m. Now we have the question of n. So you can see the answer. I will do this one. Because this is the question of a circle. I will do this question. Now the meaning of locus of a circle is Like a straight line equation came. The equation that will come here will be a circle equation. It says Z conjugate plus 2 upon Z conjugate minus 1 is 4. The real part of this is equal to 4. So you have to show that the locus of the point, that is the group of points, if we add the locus, like you remember, in childhood we used to add 1 to 2, 2 to 3, 3 to 4, 4 to 4, and at the end it used to become a lion. This is it. If you add all the points, you will get to see the equation. A plane is formed on a circle. A circle is formed on a plane. So locus means that you add a group of points, and what do you get? A circle. You add a group of points, and what do you get? A straight line. So here, we have to do the same thing. So we have done let. The best question is x plus iota y. Now we have to do x minus iota y plus 2 upon me. x minus iota y minus 1 is equal to 4. This is not 4. Its real part is 4. Now tell me, the answer to modulus will always be real. Why did you make us go around in this question? He says, real part 4. And what is modulus? Modulus has never been imaginary. What will be modulus always? Real. So, so, so, we took this as 4. So, x plus 2 minus y iota 4. Same thing again. x minus 1 minus y iota. Okay, we are on top. Let's go. I can see the circles which are not there Here it is x plus 2 whole square plus y square 4 Here it is x minus 1 whole square plus y square Now do you know the importance of this question? You knew the straight line equation Ax plus By plus C You don't know the circle equation Because this year it is going to be a complex number So how will we solve it? Let's see that First solve it as we did You will do squaring on both sides. So, this is a square plus b square plus 2ab. And here is y square. This is 16. a square plus b square minus 2ab plus y square. Okay. This is done. Now, look. This is x square plus y square plus 4x plus 4. 16x square plus 16 minus 32x plus 16y square. Bring all of them here. From 16, x will go to 15x squared. From 16, y square will go to 15y square. Right? After this, minus 32, so minus 36x came. And this came to be our 12. Now see, how is this circle equation? Let me give you a hint. We will read this in the conic section chapter. If the neighbor of x square and y square is the same, then that is the circle equation. coefficient of x square and y square is same so therefore it is the equation of circle write this in point to remember I will be very late x square's neighbor and y square's neighbor if they are same then whose equation is that? circle so what is this circle? if they are different then it will become ellipse so if they are same whether it is 1b or negative or positive, it doesn't matter if these two numbers are same x square and y square, the rest goes to hell we don't care so The neighbour of x2 and y2 should be the same. So what is a circle? If they are different, and x2 and y2 are also there, then what is that? It is an ellipse. Now what is this? We will study this in the conic section. We were told to prove the circle equation, we did that. Remember this small thing. I will explain the conic section in the chapter. Otherwise, the conic section is also a very nice one shot. You can understand by going there. There is no need for such things. There is no such urgency. Now they are saying, tell me the locus of z. Locus of z, here also I am telling you what will come, circle will come. Okay. Do not worry about pi by 6. x plus iota y minus 2 is equal to pi by 6 x plus iota y plus 2. This will come. So x minus 2 plus y iota pi by 6. You can take 6 here also. There is no problem. What did we take here? We took 6. And here came x plus 2. plus y iota. Same work. So, for now, 6 under root x minus 2 square. See, all the equations are the same. And this is pi square. And this is under root x plus 2 square plus y square. Squaring both sides. 36. This is a square plus b square minus 2ab plus y square. This is pi square. Now, pi should have been pi. Now, pi square is pi. Here comes a2 plus b2 plus 2ab and y2. Open it. 36x2 plus 36y2. Here comes 144 minus 144x. Bring them here too. Minus pi square x2 minus pi square y2 minus 4pi square minus 4pi square x. Equals to 0. Check it. Check this and this. 36 minus pi square x square. Check y as well. 36 minus pi square y square. Rest, go to the hill. Whatever came, whose equation is this? Circle's. See, what is x and y's neighbor? Same. Okay? So, this locus. Find the locus of z. So, write here, locus of z is a circle on a plane. In the last question, I told you to prove it. Here, I told you to remove it. This is the last question. You can try this easily. A2 plus B2 is given as 1. You have to find its value. Do nothing. Do rationalization. What do you have to do? Do rationalization. This is a light question. This is the last question. We are done. Practice these questions with your heart and mind. First thing, correct NCRT with examples. Third important thing, you can correct exemplar from start to end. There will be no problem. Yes, there are some polar form questions in exemplar. You will get a word, which you will call argument, which you will not have to read because it is out of the syllabus. Polar form is part of the argument, it is called theta, it is out of the syllabus. You can leave all the trigonometric questions for now because after the trigonometry, if you have done the trigonometry, then there is one more question of trigonometry, not much, there is one more because the polar form is lost. So your whole complex number is in one shot. How beneficial this lecture proved to you? Please give a note in the comment. This gives us motivation. I get it. If you see any good things in this chapter, in this lecture, please share it with your friends. Because if you do well, then you are good. Which is the next one shot? Please tell me in the comment section. I will try to make the next one. I will try to make it very soon. Now the 11th class has also started. Now 9th, 10th, 11th, 12th. The lectures of all four have started. 24th of June onwards. And now it's my turn to give a lecture here. Now July onwards, regular lecture. July, August, September, October, November. Regular lecture. Okay? So, we will meet you in the next lecture. Till then, bye-bye. Take care. Have a nice day.