in this video we're going to go over complex numbers we're going to talk about how to graph them how to calculate the absolute value we're going to work on problems on adding subtracting multiplying dividing complex numbers and even solving equations let's begin so typically you have something known as standard form which is written as a plus bi a represents the real portion of the standard or complex number so a is a real number B is the the term bi is the imaginary part of the number so let's say if we have the number 3 + 4 I how can we graph this number and also how can we calculate the absolute value of 3 + 4 I so let's make a Gra the x-axis is the real axis the Y AIS is known as the imaginary axis so the real part of the number is string so you want to travel three units to the right the imaginary part is four so you want to go up four units so the point 3 + 4 I lies right here now to calculate the absolute value of a plus bi I it's equal to the square root of a 2 + b^ 2 so basically it represents the hypotenuse of this triangle so what is 3^ 2 + 4 2 3^ 2 is 9 4^ 2 is 16 if you add them you get 25 which is five so that's the hypotenuse of the right triangle that's formed so the absolute value of 3 + 4 I is 5 now let's try another example go ahead and plot this number5 + 12 I and also go ahead and calculate the absolute value at the same time so this time we need to travel -5 units to the left that's on the real axis and on the imaginary axis we need to travel 12 units up so let's say it's somewhere in this vicinity so that's the point that we need to plot and let's make a right triangle out of it so we went five units left 12 units up and let's calculate the hypotenuse formed by this right triangle so the absolute value of a + bi I is equal to the square root of a 2 which is -5^ 2+ b^ 2 which is 12 2 - 5^2 is POS 255 and 12 2 is 14 4 if you add these two numbers you should get 169 and the square root of 169 is 13 so that's the hypotenuse of the triangle so the absolute value of5 + 12 I is 13 and this is where you plot it try this one go ahead and plot 8-5 I and find the absolute value so here's the real axis the imaginary y AIS and we need to travel eight units to the right and down 15 units so this time we're in quadrant four this is Quadrant One 2 3 and here's the fourth quadrant so the point of interest is right here that's how plot it but let's turn it into a triangle so let's travel eight units to the right and down 15 units and let's find the hypotenuse of this triangle so the absolute value of a plus bi is equal to the square < TK of 8^2 Plus + 15 2 8 * 8 is 64 and 15 * 15 is 225 -5 * -5 is also posi 225 if we add 64 and 225 that's going to give us uh 289 and the square root of 289 is 17 so the absolute value is 17 that's the length of the hypotenuse this is basically the 85 17 triangle it helps to know your special triangles so far we went over three of them there is the 345 triangle 3^2 + 4^2 = 5^2 we talked about the 52 13 triangle there's the 85 17 triangle there's also the 724 25 triangle these are the four most common right triangles that you encounter you may also see any ratios of the special triangles for example 6 8 10 works as well because that's a a ratio or multiple of the 3 4 5 triangle 10 24 26 also works and there's some other ones like 9 40 41 and 11 60 61 now if you were asked to simplify these numbers what would you do let's say if you have two num numb to simplify the sare root of 4 and the square root of4 the square root of four will give you a real number in this case two because 2 * 2 is 4 but what about the square root of4 the square root of4 is 2 I it's not just two the imaginary number I is is equal to1 so the square root of4 is an imaginary number anytime you have a negative inside an even a radical or a radical with an even index number you're going to get an imaginary number so go ahead and simplify these two numbers the square root of N and the square root of9 theare root of 9 is simply 3 the square root of 9 is the < TK of 9 * the < TK of1 the < TK of 9 is three and the < TK of 1 is I so you get 3 I simplify these numbers s < TK 25 s < TK -25 < TK 25 and negative < TK -25 so theare root of 25 is simply 5 theun of -25 is positive 5 I negative < TK 25 is going to be negative 5 and the last one is going to be negative 5 I now what about simplifying the square root of8 the square root of 18 is not a perfect square so what would you do in a situation like this well it helps to know what the perfect squares are 1 squar is 1 2 squar is 4 3 2ar is 9 4 squar is 16 5 squar is 25 6 2ar is 36 7 s is 49 all of these numbers are perfect squares you can take the square root of n and get a whole number three so you want to find out what perfect square goes into 18 the perfect square that goes into it is 9 18 is 9 * 2 and let's not forget the negative sign so we'll take out the negative 1 the square root of 9 is three we can't take the square root of two because it's not a perfect square and the square root of1 is I so the final answer is 3 radal 2 I let's try another example like that go ahead and simplify < tk50 so this is Nega * < TK 25 * < tk2 * < tk1 now 25 is a perfect square the square root of 25 is 5 and the square root of1 is I so this is the answer -5 radal 2 I try this one what is the square root of 80 16 is the highest perfect square that goes into 80 16 * 5 is 80 and the square root of 16 is four so the answer is 4un 5 times I here's one more example for you simplify < TK -72 so what perfect square goes into 72 9 goes into 72 that's a perfect square but 36 goes into 72 and 36 is larger than n so we want to write this as 36 * 2 72 / 36 is 2 that's how you can find this missing number and let's not forget the square root of netive 1 the square root of 36 is 6 and the square root of netive 1 is I so the final answer is -6 radical 2 I now let's say if you were to get a question that looks like this what is I raised to the 2011 what is that what is that even equal how would you simplify that before we can do that problem you need to understand a few things so we know that I is equal to the sare < TK of1 now what about I 2 and what about I 3 and I to the 4th power you need to know what these values are equal to as well I 2 is equal to1 if you square the square root of 1 the two and the radical will disappear giving you 1 I to the 3 is basically I 2 * I because 2 + 1 is 3 I 2 is1 so this is -1 * I or basically negative I so I 3 is equal to I and i^2 is1 the last one is I to 4th which is i^ 2 * i^ 2 and that's -1 * 1 which is POS one make sure you know these four values now let's say if you want to simplify I to the 6 power what would you do I to the 6 power is basically I to 4th power time i^ 2 the reason why you want to break it up this way is because you know that I to 4th is equal to 1 and i^2 is NE 1 so therefore I to the 6 is simply 1 so let's try another example let's say if we want to simplify I raised to the igh power now notice that a is a multiple of four so you can write it as I 4 * I 4th or simply I to 4 for a to the second power since you have two of them and 4 * 2 is 8 whenever you write it like this you need to add the exponents 4 + 4 is eight if you raise one exponent to another you need to multiply 4 * 2 is 8 now we know I to the 4th is 1 and 1^ 2 is 1 so if you have a i number with an exponent that is a multiple of eight it will always equal to one so for example I to the 12 this is one as well it's I to the 4th raised to the 3 power since 4 * 3 is 12 and 1 to the 3 power is still 1 now what about let's say I to the 15 power how can we simplify this number so so I 15 is I 12 * I 3r so one of these numbers you want it to be the highest multiple of four that's closest to 15 multiples of four are 4 8 12 and 16 16 is too much so you want to choose 12 and 15 minus 12 gives you three so you want to break it up this way and I to 12 is I to 4th raised to the thir power which is 1 to the 3 power and we know that I Cub is equal to I so the final answer is 1 * I or simply negative I now for those of you who may want to know another way or systematic way of getting these two numbers especially if this number becomes very large here's what you could do take 15 and divide it by four if you type this in your calculator you should get 3.75 now take the whole number portion of this uh decimal number and multiply by four 3 * 4 is 12 that gives you this number now the decimal portion of the number multiply that by 4 so 75 * 4 gives you three which is the other remaining number that's how you could find these two numbers if you ever have difficulty seeing what it is let's try another example simplify this one I to the 29th so this is I to 28 * I and I to 28 is basically I to 4th raised to the 7 since 4 * 7 is 28 and this is going to be I to 4th is 1 and 1 to the 7th power is 1 so 1 * I is simply I so I to the 29th is the same as I so one way to get these numbers 28 and 1 take 29 / by 4 29 ID 4 is 7.25 so then what you want to do is multiply 7 by 4 which gives you 28 that's the uh first number and then the decimal portion of the number 0.25 multiply that by four you should get one which is the the number what about I raised to the 62 this is I to 60 * i^ 2 and I to the 60 is basically I to 4th raised to the 15 power I to 4th is 1 i^ 2 is 1 1 rais to anything is 1 so so 1 * 1 is1 let's try one more problem I raised to the 201 so let's work on this one so what two numbers should we use so let's use that process that we were dealing with earlier 2011 / 4 this is equal to uh 50.2 now if you multiply 4 by 50 this will give you 200 and 4 * .25 is equal to 1 so this is going to be 200 and 1 now 200 / 4 is 50 so you want to write it as I to 4th raised to the 50th power time I to the first Power and I to the 4th is 1 and 1 raised to the 50th is simply one so it's 1 time I which means the final answer is I so for these type of problems there's only four possible answers it's either I or i^2 which is 1 I Cub which is negative I or I to 4th which is 1 so it's always going to simplify to one of these four values our next topic is adding and subtracting complex numbers so let's say if you have a problem that looks like this 5 + 2 I plus 3 + 7 I what would you do so in a situation like this all you need to do is combine like terms the real numbers are five and three so we can add those two numbers 5 + 3 is 8 the imaginary numbers 2 I and 7 I if we add them that is going to give us 9 I and that's all you need to do for this particular problem so let's try another example go ahead and add 7 + 3 I + 6 + 5 I so 7 + 6 is 13 3 I + 5 I is 8 I try this one 4 + 8 i - 3 - 5 I now we only have a one in front of the first parenthesis so if we multiply everything by one it's going to stay as 4 + 8 I now here we need to distribute the negative sign so it's going to be -3 + 5 I so be careful when you have a negative because some people may forget to change this sign so now let's add the real numbers 4 - 3 is 1 8 I + 5 I is 39 So This Is The Answer 1 + 39 now what is the absolute value of 1 + 39 so don't forget the absolute value is going to be the square root of a 2 + b^ 2 or 1 2 + 13 2 13 squared is 169 and if you add one to that you're going to get the square root of 170 now 170 is divisible by 10 this is the square root of 7 and aun of 10 there's no perfect square that goes into 170 so this is the final answer go ahead and simplify this problem 7 * 4 + 3 I minus 5 * 2 - 6 I so first let's distribute 7 to 4 + 3 I so 7 * 4 is 28 and 7 * 3 I is 21 I now let's distribute the five5 * 2 is -10 and5 * -6 I is POS 3i so now let's combine like terms 28 - 10 is 18 and 21 I + 30 I is 51 I so this is the answer in standard form or in a plus bi form try this four plus the square < TK of -25 + 3us < tk81 go ahead and simplify this expression so this is 4 plus we know the square root of 25 is 5 so the square root of - 25 is 5i and the square root of 81 is 9 I so we can add 4 + 3 which will give us 7 and 5 i - 9 I is -4 I so this is the answer try this one 7 minus the < TK of 9us -4 + theare < TK of -36 so this is going to be 7 minus theare < TK of9 is 9 I and then if we distribute the negative sign this is going to be + 4 and then minus the square otk of -36 6 I so 7 + 4 is 11 and 9 i - 6 I is -5 I so this is it what about this one what is 8 I multiplied by 4 I what's the answer how would you simplify it 8 * 4 is 32 I * I is i^ 2 and as you recall i^ 2 is1 so the final answer is -32 try this one what is 5 I raised to the second power so 5^ 2 is 25 * i^ 2 so this is going to be 25 * -1 which is -25 now what about this what's 3 itimes 5 I * -7 I so 3 * -7 is -21 and I * I * I is basically I to the 3 power 21 * 5 is 105 and I to the 3 is negative I so the final answer is postive 105 * I simplify this expression what is 5 itip 4 - 2 I go ahead and work on this example so let's distribute 5 I * 4 is 20 I and 5 I * -2 I 5 * -2 is -10 I * I is i^ 2 so we could simplify the i^ 2 part i^ 2 is1 and -10 * -1 is positive1 now we need to put it in standard form that is in a plus bi form so we need to reverse the two numbers therefore the final answer is 10 + 20 I in standard form try this one what is 5us 3 I * 4 + 7 I in this example we need to foil so what's 5 * 4 5 * 4 is 20 and then 5 * 7 I is 35 I -3 I * 4 is -12 I and -3 I * 7 I is -21 i^ 2 so we can combine 35 I and 12 I 35 - 12 is 23 -21 i^2 that's -21 * 1 which is posi 21 + 20 so that's 41 so the final answer in standard form is 41 + 23 I go ahead and work on this example ulti 6 - 5 I * 3 + 8 I so go ahead and foil 6 * 3 is 18 and 6 * 8 I is 48 I5 I * 3 is-5 I and finally 5 I * 8 I is -40 i^ 2 so let's combine 48 I and 59 so 48 - 15 that is equal to 33 or in this case 33 I and i^2 is 1 so this is going to be 184 * -1 is + 40 and 18 + 40 is 58 so the final answer is 58 + 33 I now what if you were to see a problem like this 4 + 5 i s what would you do to simplify in standard form so what this means is that you have two 4 + 5 I terms multiply to each other so you want to write it out like this expand it and then four so 4 * 4 is equal to 16 and 4 * 5 I is 20 I 5 I * 4 is also 20 I and 5 I * 5 I is 25 i^2 so let's combine the two middle terms so that's going to be 20 + 20 is 40 and 25 i^ 2 is - 255 so what is 16 - 25 16 - 25 is9 so the final answer is9 + 40 I here's one for you go ahead and simplify this expression in standard form 5 - 3 I raised to the 3 power so we need to expand it or write it three times so let's foil two two of these binomials at one time so let's start with those two what's 5 * 5 5 * 5 is 25 and then we have 5 * -3 I which is -5 I and then -3 I * 5 which which is also -5 I and finally -3 I * -3 I which is posi 9 i s and we still have another one on the outside so let's simplify this expression -5 I and5 I adds up to -30 I we still have the 25 and 9 i^2 is9 so now we can combine 25 - 9 which is 16 so we have 16 - 30 I * 5 - 3 I so at this point we need to foil these two expressions so 16 * 5 is 80 16 * -3 I is -48 I and uh -30 I * 5 is 50 I and finally -30 I * -3 I is positive 90 i^2 so I believe my math is correct hopefully I didn't miss anything so once again let's add the two middle terms what's -48 plus- 15 these two they're going to add to - 198 * I and 90 i^2 is basically 90 80 + 90 is -10 so the final answer is -10 - 198 * I now you need to know what's going to happen when you take a complex number and multiply it by its conjugant so consider the complex number 3 + 4 I the conjugate of this number is 3 - 4 I it has the same A and B value the only difference is this is a positive B value and this is a negative B value when you multiply a number by its conjugate you're going to get two terms initially and ultimately it's going to simplify to a real number the imaginary numbers will cancel so here's the quick way to get the answer it's going to be 3 * 3 which is 9 and 4 I * -4 I which is -16 i^ 2 and i^ 2 is 1 so that's + 16 which is 25 that's the fast way but I'm going to show it to you by foiling the entire problem so 3 * 3 is 9 3 * -4 I is -12 I 4 I * 3 is POS 12 I and finally 4 I * -4 I is -16 i s so notice that the middle terms cancel anytime you multiply a complex number by its conjugate and i^ 2 is 1 so -16 i^ 2 is -16 * 1 which is pos6 and so the final answer is 25 so notice that this no more imaginary numbers we just get a real number so let's try this example 5 - 2 I multipli its conjugate 5 + 2 I so for this particular example we just need to multiply the first two terms 5 * 5 which is 25 and the last two terms -2 I * 2 I which is -4 i^ 2 so this is-4 * -1 which is +4 and 25 + 4 is 29 now let's say if you would to see a question like this what is 3 * 7 I SAR compared to 3 + 7 i^ 2 so what's the difference between the two and how would affect the way you would solve it so for the one on top notice that the three is multiplied by the 7 I so therefore this is equivalent to 3^ 2 time 7 i^2 you can distribute the exponent below you can't do that you can't say this is 3^2 + 7 i s it doesn't work that way for the one on the bottom you need to foil it you need to expand it first as 3 + 7 I * 3 + 7 I and then for so if you have an addition or a subtraction sign between the three and the s i you have to foil it if you have a multiplication or division sign you can distribute the two you don't have to foil it make sure you understand the difference so for the example above 3 S is 9 7 s is 49 9 and i^ 2 is just i^ 2 so 9 * 49 is uh 441 * i^ 2 which is - 441 now for the example below we need to foral 3 * 3 is 9 3 * 7 I is 63 3 I plus I messed up there 3 * 7 is not 63 3 * 7 is uh 21 and we're going to get another 21 I and finally 7 I * 7 I is 49 i^ 2 so we have 9 + 42 i - 49 9 - 49 9 is -4 now what about dividing complex numbers let's say if we have 4 + 3 I / 5 - 2 I what would you do to simplify this expression if you're dividing complex numbers focus on the denominator which is 5 - 2 I multiply the top and the bottom of the fraction by the conjugate of the denominator so if you see a minus sign make sure this is plus whatever you do to the bottom you must also do to the top in order that the fraction maintain its value so let's foil 4 * 5 is 20 4 * 2 I is is 8 I 3 I * 5 that's 15 I 3 I * 2 I is + 6 i^ 2 now because these two are conjugates we only need to multiply the first and the last term the two middle terms will cancel 5 * 5 is 25 and -2 I * 2 I is -4 i^ 2 so let's combine 8 I and 15 I 8 + 15 is 23 and 6 i^2 is -6 -4 i^ 2 is going to be + 4 so 20 - 6 is 14 and 25 5 + 4 is 29 now we need to put this in standard form so let's separate this fraction into two smaller fractions let's divide both numbers by 29 so this is 14 over 29 plus 23 over 29 time I so it's now in a + bi form or standard form try this one divide 8 by 6 + I so just like the last example we're going to multiply the top and the Bottom by the conjugate of the denominator so let's distribute 8 to 6 I 8 * 6 is 48 and 8 * I is simply 8 I on the bottom we need to foil but since they're conjugants we could just multiply the first two 6 * 6 is 36 and the last two I * I which is i^ 2 i^ 2 is -1 so negative i^2 must be positive 1 so this is equal to 48 - 8 I / 37 and now let's separate it into two smaller fractions so the final answer is 48 / by 37 - 8 over 37 * I so this is the answer in standard form so for the sake of practice try this one 7 + 2 I / 3 - I pause the video and work on this example so let's multiply by the conjugate of the denominator so on topl this foil 7 * 3 is 21 7 * I that's 7 I 2 I * 3 6 I and 2 I * I which is 2 i^ 2 on the bottom the first two 3 * 3 is 9 and the last two I * I I is I 2 so let's add 7 I and 6 I that's going to be 13 I 2 i^2 is -2 i^ 2 is + 1 21 - 2 is 19 9 + 1 is 10 so the final answer is 19 / 10 + 13 / 10 * I what if you have a complex number and you wish to divide it by an imaging number what would you do in this problem so to simplify this expression your goal is to get rid of the imaginary number to do that multiply the top and the Bottom by I if you can turn it into I2 or I to the 4th power you can get rid of the imaginary number so on top let's distribute I 5 * I is 5 I and -2 I * I is -2 i^ 2 on the bottom I * I is simply i^ 2 so we have 5 I -2 i^ 2 is pos2 i^ 2 is 1 so this is 5 I /1 which is 5 I and 2 /1 is -2 so this is equal to -2 - 5 I so let's work on some more examples let's try 3 + 2 I divid 7 I so for this example we need to multiply the top and the Bottom by I so on top if we distribute I it's going to be 3 I + 2 i^ 2 and on the bottom we're going to have 7 i^ 2 so 2 i^ 2 is -2 7 i^ 2 is -7 so I'm going to rewrite it as -2 + 3 I so it's going to be in standard form so -2 / 7 or 7 that's pos2 over 7 and and then 3 I / -7 is3 / 7 * I so this is the answer in standard form go ahead and simplify the expression so this one it looks weird but it's not that bad we just got to multiply the top and the Bottom by I so on top it's going to be 9 I on the bottom I to the 4th and remember I to the 4th is one so the final answer is simply 9 I now what if you were to see a problem that looks like this 5 - 2 I / 2 + 3 I 2 what would you do to simplify so before we can multiply by the conjugate or by I we need to simplify this expression we need to put it in a plus bi form so let's foil 2 + 3 I time another 2 + 3 I so 2 * 2 is 4 2 * 3 I is 6 I and 3 I * 2 is also 6 I and then 3 I * 3 I is 9 i^ 2 on the top everything's going to be the same 6 I + 6 I is 12 I and 9 i^ 2 is 9 so now we can combine 4 and 9 which is5 so it's 5 + 12 I on the bottom so now at this point we can multiply the top and the Bottom by the conjugate of the denominator that is5 - 12 I so on top let's distribute 5 * 5 is - 255 5 * -12 I that's -60 I and -2 I * -5 is positive 10 I and finally -2 I * -12 I is positive 24 I2 so hopefully I didn't miss any negative signs or anything like that it's very easy to make a mistake things happen but I'm just double checking my work so everything looks good so far on the bottom since they're conjugates of each other we could just multiply the first two and the last two5 * 5 is POS 255 12 I * -12 I is - 144 i^ 2 so now let's combine -60 and 10 I -60 + 10 is50 24 i^ 2 is -4 and - 144 i^ 2 is+ 144 so now let's add 25 and4 which is - 49 25 + 144 is 169 so the final answer for this problem is -49 / 169 - 50 over 169 times on so as you can see whatever expression you have that's a complex number you can always put it in standard form there's always some technique that you can employ to put it in a plus bi form so let's say if you have two equations x^2 - 36 is = 0 and also x^2 + 36 is equal 0 what would you do to solve for x in the first example we can Factor it using the difference of squares method the square root of x^2 is x the square root of 36 is 6 one will be positive and the other will be negative so therefore X is equal to -6 and POS 6 so that's what you could do if you have the difference of perfect squares if you have the sum of perfect squares it's going to be x + 6 I * x - 6 I so therefore X is equal to Plus or - 6 I notice that these two are conjugates of each other so you could check your answer by foing so x * X is x^2 and 6 I * -6 I is - 36 i^ 2 which is POS 36 so whenever you're factoring a sum of perfect squares you're going to get imaginary numbers if it's a difference of perfect squares you're going to get real numbers so let's try some more examples solve for x so let's try this one 3x^2 + 48 is equal to0 let's put it in factor form and then we'll solve for x so we can take out the GCF which is 3 3x^2 / 3 is x s 48 / 3 is 16 so notice that we have the sum of perfect squares X squ is a perfect square 16 is um is a perfect square because you can take the square root of it the square root of x^2 is x the square root of 16 is 4 but we're going to have 4 I and 4 I one is going to be positive the other will be negative so therefore X is equal to POS 4 I and -4 I so let's try another one 4x^2 + 100 is equal to Z go ahead and work on that example so let's take out the greatest common factor 4x^2 / 4 is x^2 100 ID 4 is 25 so to factor it it's going to be x + 5 I since the < TK of 25 is 5 and x - 5i so to solve for x let's set each factor equal to zero so in this example we need to subtract both sides by 5 I here we need to add by 5 I so we can see that X is equal to 5 I and positive 5 I consider this one -9 x^2 - 100 is equal to zero what would you do to solve this particular example what I would recommend is taking out the GCF which is 1 -9 x^2 /1 is 9x2 -00 /1 is plus 100 so notice that we have the sum of perfect squares 9 and 100 are perfect squares the square root of 9x2 is 3x the square root of 100 is 10 but it's going to be 10 I since we know we have the sum of perfect squares so one is going to be positive and the other is going to be negative so we could write three equations let's set I mean two equations let's set each factor equal to zero so 3x plus 10 I is equal to 0 and 3x - 10 I is equal to 0 so therefore X is going to be -10 I / 3 and it's also going to be positive 10 over 3 * I let's try this one x^2 + 1/ 9 is equal to0 so here we have a fraction go ahead and solve for x the square root of x^2 is X the < TK of 1/9 is 1 / 3 now let's not forget to put the imaginary numbers so we're going to have a positive and a negative sign so we could see that X is equal to - 1 3 * I and positive 1 3 * I and that's all you got to do for that particular example so how would you solve this particular equation 3x^2 + 4x + 7 now typically you would try to see if you can Factor this expression since we have a trinomial 3 * 7 is 21 if we could find two numbers that multiply to 21 but that add to the middle term four then this expression is factorable the only factors of 21 are 1 and 21 and 3 and 7 which none of these add up to four we also have -1 and - 211 and3 and7 now that doesn't add to U 21 unless we had like -3 and posi 7 that would add to four but -3 * 7 doesn't multiply to positive 21 it multiplies to - 21 so this expression is not factorable so therefore the only way to solve it is either to complete the square or to use the quadratic formula this equation is in standard form ax^2 + BX plus C so a is 3 B is 4 C is 7 so let's use the quadratic formula X is equal to B plus or minus Square < TK b^ 2 - 4 a c / 2 a so B is 4 which means that b^ 2 4 2 is 16 a is 3 C is 7 / 2 * a or 2 * 3 so this is going to be -4 plus or minus 16 now 3 * 7 is 21 and 21 * 4 is 84 divided by six so now what is 16 - 84 16 - 84 is -68 can we simplify the square root of 68 it turns out that we can the square root of 68 is basically 4 the < TK of 4 * the < TK of 17 4 * 17 is 168 and let's not forget the negative 1 the sare root of 4 is 2 and the square < TK of1 is I so this is what we have at this point so there's two answers but let's separate it into two fractions so it's -4 / 6 plus orus 2 R 17 / 6 * I so we can reduce 4 over 6 if we divide both numbers by two it's going to be -2 over 3 and 2 over 6 we can reduce it 6 / 2 is 3 but it's going to be on the bottom so it's the sare < TK of 17 / 3 * I so the two answers are -2 over3 + < tk7 over3 * I that's in a + b i form and the other answer is -2 over3 minus < tk7 over3 * I let's try another example try this one 2x^2 - 3x + 9 go ahead and solve for x so let's use the quadratic equation one more time it's negative B plus or minus sare < TK b ^ 2 - 4 a c ID 2 a so in this problem B is -3 so b^2 or -3 2 that's going to be 9 a is 2 and C is 9 / 2 a or 2 * 2 so negative * -3 is posi3 4 * 2 is 8 8 * 7 I mean 8 * 9 is 72 and 2 * 2 is 4 so what is n 9 - 72 9 - 72 is uh -63 so notice that we could simplify otk 63 otk 63 63 is basically 9 * 7 and let's not forget the sare root of NE 1 so the square root of 9 is three and the square Ro T of1 is I so this is what we now have so let's separate it into two fractions so this is going to be 3 over 4 plus or minus 3 Ral 7 / 4 * I and so the two answers are 3 over4 plus 3un 7 over 4 * I and the second answer is 3 over4 minus 3un 7 ID 4 * I so these are the two imaginary Solutions now what if you were to see an equation that looks like this 2x * x - 3 is equal to 5 or rather let's say 5 what would you do to solve for x so notice that this is a quadratic equation but not in standard form so we got to put it in standard form before we can use the quadratic formula so 2x * X is 2x^2 2x * -3 is -6x and let's add five to both sides so it's 2x^2 - 6 x + 5 which is equal to zero so now we can use the quadratic formula now that it's in standard form so now B is -6 and then we have b^2 or -6 s minus 4 and a is 2 C is five / 2 a or 2 * 2 so this is going to be 6 Plus orus Square < TK -6 2 is 36 2 * 5 is 10 and 10 * -4 is -40 and 2 * 2 is 4 so let's make some space so 36 - 40 is-4 and the square root of -4 the square root of 4 is 2 so the square root of4 is 2 I so let's separate it into two fractions so this is going to be 6 over 4 plus or- 2 over 4 * I 6 over 4 reduces to 3 over2 if he divide the top and bottom by two and 2 over 4 reduces to 1 over 2 * I so therefore we have two answers 3 over2 + 12 I and the second answer 3 over2 minus 12 I and that's it for this problem now what if you were to see an equation like this 6X + 2 I is equal to 18 + 12 y i so there's two variables X and Y what can you do to solve for x and y now you need to keep in mind that these complex numbers have two components the real component and the imaginary component the real component on the left side is 6X because it doesn't have an I attached to it the real component on the right side is 18 so therefore we could say that 6x must be equal to 18 now the imaginary component contains an i the imaginary component on the left side is 2 I and on the right side is 12 Yi so therefore we could say that 2 I is equal to 12 Yi on the left we could divide by six so we could say x is equal to 3 and on the right side to solve for y we could divide both sides by 12 I to get y by itself so on the right 12 I will cancel leaving us with Y on the left the I variable will cancel and it's 2 over 12 if you divide it backwards 12 / 2 is 6 so 2 over 12 must be 1 over 6 So This Is The Answer X is equal to 3 and Y is equal to 1 over 6 try this one 3x + 4 I is = to 15 + 16 Yi so let's set 3x equal to 15 since both of those terms are real numbers and let's set the imaginary Parts equal to each other so 4 I is equal to 16 y i so let's divide by 3 15 / 3 is 5 so X is 5 and for the second equation let's divide by 16 I so on the right side all we have is just y on the left side the I variables cancel 16 / 4 is 4 so 4 16 must be 1 over 4 and so those are the answers try this one what is x + 4 I * x - 4 I and let's say that's equal to 41 what is the value of x so we have two conjugates multiply to each other so we can multiply the first and last terms x * X is x^2 and 4 I * -4 I is -16 i^ 2 and since i^ 2 is -1 -16 i^ 2 is + 16 so now what we need to do is subtract both sides by 16 41 - 16 is 25 so we could take the square root of 25 therefore X is going to be plus or minus 5 so that's the answer to the problem now let's say if you're given the imaginary solutions to an equation and you're asked to write that equation what would you do so first you want to write it in factored form if your answer is 3 I you want to write x - 3 I if it's- 3 I write x + 3 I of course this is probably equal to zero you may or may not need that zero and then you want to foil since these are conjugates of each other we could just multiply the first and the last two terms so it's going to be x^2 time -9 i^ 2 which is x^2 + 9 so that's how you can find the equation if you're given the solutions so let's try another example so let's say if you're given just one solution for I what's the equation now for a quadratic equation imaginary numbers they come in pairs so if you have 4 I you must also have the conjugate 4 I so therefore we can write x - 4 I * x + 4 I and we know it's going to be x^2 - 16 i^ 2 which is x^2 + 16 so basically you're working backwards now let's say if you're given one of two imaginary Solutions let's say 4 + 3 I and you want to write the equation first you need to find the other imaginary solution which is going to be the conjugate for minus 3 I so how can we write these two numbers in factored form so here's what you need to do notice that you have a plus4 change the sign of four and it's going to be minus 4 and then change the sign of plus 3 I it's going to be minus 3 I do the same thing for the other side so we have a posi 4 change it to ne4 -3 I change it to positive 3 I and so now we need to foil so there's an easy way and there's a long way we're going to do it the Long Way first and then we'll do it the easy way so you'll see why it works here we have three terms three terms initially when we foil it we should get nine terms and then we'll simplify so x * X is x^2 x * -4 -4x x * 3 I that's going to be 3x I and then -4 * X -4x -4 * -4 is [Music] 16 and -4 * 3 I is -12 I and then we have -3 I * X which is -3x I and then -3 I * * -4 that's pos2 I and finally -3 I * 3 I is 9 i^ 2 so now what terms cancel we can cancel the 3x I and the -3x I we can also cancel the -12 I and a 12 I so what do we have left over so we have X2 and these two combined to form -8x + 16 - 9 i^ 2 -9 i^ 2 is POS 9 and 16 + 9 is 25 so this is the quadratic equation x^2 - 8x + 25 now let's start back from our original factored expression which was x - 4 + 3 I and x + 4 I mean minus 4 but - 3 I so our goal is to get this answer x^2 - 8x + 25 so another way in which you can do it is you're going to multiply x - 4 * x - 4 which is x - 4^ 2 and then you're going to multiply these two 3 I * -3 I so when you foil x - 4^ 2 because it's a perfect square it's going to be x^2 and it's going to be -4x +4x and -4 2 is going to be 164 * 4 and 3 I * 3 I is 9 i^ 2 so this is going to be x^2 - 8x + 16 + 9 which turns into this so that's how you can do it the fast way let's try another example so given these two solutions go ahead and write the quadratic equation so let's write it in factored form since we have positive 5 it's going to be x - 5 + 2 I and x - 5 - 2 I so this is the same as x - 5^ 2 + 2 I * -2 I so x - 5^ 2 that's going to be x - 5 * x - 5 which is x 2 - 5x - 5x and5 * 5 is 25 and 2 I * -2 I is -4 i^ 2 so this is going to be -10x + 25 -4 i^ 2 is + 4 so the final answer is x^2 - 10 x + 29 so this is it now what if you were to see something like this x equal to the < TK of 7 + 3 I what's the other answer the other answer is going to beunk 7 but minus 3 I so let's go ahead and convert this into a quadratic equation using both techniques so this is going to be since we have a positive in front of the radical 7 it's going to be xus otk 7 - 3 I and x minus otk 7 but plus 3 I so let's foil it the long way x * X is x^2 x * < 7 that's otk 7x and plus 3x I and thunk 7 * X and then this is going to beunk 7 * 7 which is simply -7 because that's the that's actually positive 7 the two negative signs cancel otk 7 * otk 7 is the < TK of 49 which is 7 andun 7 * 3 I that's going to be -3 < tk7 I -3 I * X is -3x i-3 I * > 73 well it's going to be positive now posi 3un 7 * I you got to be careful it's very easy to like make a mistake -3 I * 3 I is 9 i^ 2 so now let's cancel the terms that can be canceled that's 3x I and also uh 3un 7 I so we're left with X2 these two we could combine that's going to be 1 radal 7 + 1 radal 7 is 2 radal 7 so this is going to be -2 radal 7 * X and then + 7 -9 i^ 2 is going to be + 9 so the final answer is x^2 - 2un 7 * x + 16 now let's get the same answer using the other technique so in factored form it was x - otk 7 - 3 I * x - > 7 + 3 I so this is equivalent to being x - < TK 7^ 2 and then plus 3 I * -3 I so x - > 7 * x - > 7 that's going to be x^2 - TK 7X - another < TK 7x andun 7 * 7 is POS 7 and 3 I * -3 I is 9 i^ 2 so we can see that these two they're going to add to -2k 7x and this is going to be 7 9 i^ s is + 9 which these two will become 16 so you get the same answer now let's say if you're given a quadratic equation in standard form ax^2 + BX + C and you want to find the sum and the product of the roots in a quadratic equation there's usually two Roots two real solutions or two imaginary Solutions or one real Solution that's the same but two answers how can we find the sum and the product of the roots if we're given the quadratic equation the equation is this the sum is equal to B / a and the product is C / a so let's try an example let's say if we have the quadratic equation x^2 + 36 = what is the sum and the product of the roots in this equation so this is 1 x^2 + 0x + 36 which is equivalent to x^2 + 36 so we can see that b is zero so the sum is going to be0 / a which a is one so the sum is zero the product is going to be C over a c is 36 so it's 36 over 1 which is 36 now let's prove it so the equation x^2 + 36 can be factored as x + 6 I * x - 6 I therefore the roots are - 6 I and positive 6 I so if we are looking for the sum of the roots the sum of the roots is just adding or two answers - 6 I + 6 I adds up to zero now the product we just got to multiply it -6 I * 6 I will give us the product which is -36 i^ 2 and since i^2 is1 this is going to be positive 36 let's try another example x^2 - 4x + 29 so given this quadratic equation with three terms in other words a trinomial find the sum and the product of the roots so the sum of the roots is Nega B / a the product is simply C / a so use the equation and also find the solutions and confirm the answer so let's use the equation first B is-4 A is 1 so the sum is four which I'm going to write it right here the product is C over a where C is 29 a is 1 so the product is 29 so now let's find the solutions and let's confirm that answer so let's solve for x using the quadratic equation B plus orus < TK B ^2 - 4 a c / 2 a so B is -4 and b^2 -4 2-4 * -4 is 16 a is 1 C is 29 / 2 a or 2 * 1 netive * -4 is pos4 4 * 29 that's going to be 116 / 2 16 - 116 is NE 100 now the square root of 100 we know it's 10 so the square Ro T of 100 must be 10 * I which is a nice number so now let's separate into two fractions that's going to be 4 over 2 plus orus 10 / 2 and 4 / 2 is two 10/ 2 is 5 so there's two answers 2 + 5 I and 2 - 5 I so these are the two complex Solutions now that we have the solutions we can confirm if the sum is four and if the product is 29 so let's go ahead and do that to find a sum we need to add the two solutions 2 + 5 I plus 2 - 5 I so 2 + 2 is 4 5 I + 5i they cancel it's zero so the sum is indeed four now let's find the product of the two solutions so the product is going to be 2 + 5 I multiplied 2 - 5 I now because these are conjugates of each other we can multiply the first two and the last two 2 * 2 is 4 5 I * 5 I is - 25 i^ 2 -25 i^ 2 is + 25 since i^ 2 is 1 and 4 + 25 is 29