hi hello Namaste W welcome you all okay guys already High weightage series he Infinity Target 99 plus percent number ofam tricks shortcuts hi aant hi Emma prano vishu uh tja uh ni Sai na hi hi hi very very good evening so session start start let us move into the session okay now okay now right so let us move into the session com Quadra equations and 18 to I and uh differential equations along with areas LC number of topics set relations functions sequence series PNC probability stats matrices and determin okay right so complex numbers comple numbers reality real numbers 1 2 3 4 root2 root3 root4 1 by 3 1 by 5 real numbers but comp sub so 2 + 3 I so X number y numberal < tk3 minus I into 1 by 63 number 1 by 6 number a i 6 so junk3 I so the real numbers are the subset of the complex numbers so num comp imaginary numers all the real numbers are comp numbers i² = a is aiti won't become Maybe real imaginary numbers so squaring on both Sid square root on both square root of x = root ofus so X IM IM comp important topic PhD master of science sorus root roare root1 into 3 square square root of 3 square and 3 square root of minus1 so this is 3 into I square of sir I uploaded 2023 certificate instead of 202 cic congratulations congratulations okay right soab I andare root of minus1 I Square - oneing both iare = root ofus square whole square sare root canc I right I square-1 into IUS I I power 4 I sare sare I I sare sare which is +1 so I sare and-1 I i- I I power 4 1 so b 1 I CU and- I I power 4 and 1 I 5 i i 6us i 7 i i i 9 I I power 10- I power 11us I I power 12 iusus IUS plus plus I examp 200 I 200 next I power 2025 when will class okay right I power 200 I 200er shortcut 50 so I anything so third one I power 22 so 2 4 2 are 8 and 22 4 5 are 20 and 22 very simple question number next property sum of four consecutive I poers sum of four consecutive I Powers is always zero and I power n + I power n + 1 + I I power n + 2 + I power n + 3 can be any integer I power 5 + I power 6 + I power 7 + I Power 8 I and I I 6 and I sare -1 i^ 7- I I 8 1 so total Nal I powerus 3+ I power -2 + I power -1 + I power 0 yes po I IUS by momat i sare i iusus i 1 by IUS I 1 by icial root but 1 by root minus1 minus root minus1 meaning okay important for example question summation IAL 1 to 100 uh 100 I power R question number one question number two summation IAL 0 to 100 I power r value third one summation I = to - 2 to4 i r question fourth question summation I = -1 to 100 1 by I power R different different questions okay fth one sum r r isal to 1 to 100 1 to 100 I power so Su sum of the terms okay ADV G government funded 1 I 2 I I I power 5 I power 6 I power 7 I Power 8 0 ter that is find number of terms number terms 100 terms 100 terms what is theer 4 25 are 100 Z no terms are left all Zer all zer number term number of terms number of terms and so 100 one term left reminder indicates number of terms left so Remer I very number of terms number of terms r valus - 1 0 1 2 3 and so on4 r value minus 2us one what is the remainer S4 Z 0 I power minus 2 I power -2 + I power minus1 I power before consec terms value Z 0us IUS I is the correct answer very good superus one to start 0 1 2 3 and so on 100 I power minus of minus1 + 1 plus r power I power so this is I + 1 1 plus I is the correct answer simple I and so I 100 2 + 3 plus and so on 100 so sum of 100 natural numbers 100 into1 by 2 so I power 50 into 19 50 into1 19 okay so 50 into1 number 5 so uh 40 48 48 25 24 so IUS so product R = 100 I rus1 November M crash course please yes Kumar m now right first question summation k equal to 0 to 200 I power K product P equal 1 to 50 i p product so I power I power P value 1 + 2+ 3 plus and so on up to 50 so I power n into n + 1 by 2 25 25 into 51 25 into 51 25 so 1275 so 1275 4 35 35 I I 1 + IUS I into minus1 y so XA y 1us 0a 1ms clear Andy right so I added flash class on 31st October 31st October just right the smallest positive integer for which this equal this thisal 1 + I 2 I 2 a + b² + 2 a power n a² + b² - 2 A B whole power n 1 sare 1 I sare -1 + 2 i n 1 I sare -1 - 2 i^ N 1 - 1 g 1 - 1 gone this is 2^ n into I power n is equal to the - 2^ n into I power N9 okay now so I N I N cancel 2 N = - 2 n must be number so number sorry one even is POS right let us move further soal 2024 in the set Al a b a = 42 and X iy then the value of m+ x + y a + 5al 42 42 37 so Val 42 4 Val but B value 9 42 45 which isus 3 a valus numb 1 2 3 4 5 6 s 1 to8 m 8 into 1 - I power n factorial equal x = 1 2 m 1us summation n = 1 m 8 i n fact 8us Nal 1 I 1 factorial I power 1 plus Nal 2 I 2 factori 2 I 3 factal 24 4 factori 120 so 8us i i s i values Zer sorry sorry 24 i 4 6 I 1+ 1 + 1+ 1+ 1 minus I minus orus + 2 + 5 and + 7 15 I I x y x + y + m 15 - 1 + 8 which is 22 oh m + x + y 8us I can - I 5 - I so this is nothing but 5 - I 12 is the correct answer 12 is the correct answer yes 12 is only the correct answer comple numb so comp num very very very important basic algebra number one basic algebra basic algebra of complex numbers Addus Z1 into Z2 Z1 by Z2 very very important conjugate number two conjugate number three modulus arguments geometry of the complex numbers deleted delet conjugate important conjugate 2 imaginary ofus by 2 the given complex number is real then find out the values of theta the given complex number is purely imaginary then find out the values of theta if Z Plus z bar is this much then what is the value of this question so conjugates Maxx important conjugate very very very important the modul very very very important the basics very very important the geometry very very important s logi of a complex number argument of a comp number square root of comp number question modulus next conjugate next Basics very very Basics geometry s s basic form Maxim right soal I X plus I comple number real imaginary imaginary part okay IM part real IM IM plus I Y so real of plus I into imaginary of comple number comp number imal x + i y x i imaginary number one two equal Z + z bar by 2 x 2 important Point okay imag 2 I into y equal imagin y equal important point very important IM imagus 2 i imag yal z by 2 IUS Z baral 2 I 2 IM imagin real of0 meaning the third one it is z is purely real Pur number J is purely imaginary purely imaginary IM imag imag IAL ah that implies zal X complex numberal imagary IM is purely imaginary imag y x x x i purely imaginary is purely imaginary imaginary real purely imaginary real zero imaginary imaginary Z real IM imag so purely imaginary part of right purely imaginary imagely imag imag imag albra of the comp number so what is the algebra of the complex number basic comp comp number X2 X1 X2 I y1 into I Y2 I sare y1 Y2 I Square minus one so minus y1 Y2 X1 into I Y2 y i y1 into X2 I X1 Y2 + y1 X2 X1 Y2 plus y1 so comp numb equ what real part equal and imaginary partal imaginary x - 2 + uh 3 I is equal to uh uh is equal to 4 + a a number imaginary part equal imaginary part and x - 2 real 4 real x - 2 = 4 and x = 6 AAL 3 36 + 27 so which is nothing but 13 163 so mat chapter INF book okay right very good afternoon for example examp 2 + I by 1- I 2 + I by 1- i a i a b compat 1 a a Aus b s and I square and minus 1us into minus + 1 in the 1 + 1 2 numer 2us 1 1 1 I 2 i+ 3 I so 1 + 3 I A 1 by 2 B 3 by 1 by 2 3 by2 4 by 2 4 by 2 2 so a plus b answer two is the correct answer very good very good professional sir doubt sir please 29th January for two nonzero complex number Z1 and Z2 real part of Z1 into Z2 0 real part of Z1 + Z2 0 then which of the following is are are possible options X1 + I y1 Z2 X2 + I Y2 so what is into X1 X2 - y1 Y2 real part plus I into imaginary part X1 Y2 y12 real part of this real part of Z 1 Z2 real part Z X1 X2 minus y1 y X12 y1 what is2 X1 + X2 I into y1 + y X1 + X2 = 0 second IM imag of imagin of y1 imagin of2 Y2 y1 and y1 X2 = -1 y1 into Y2 X1 into X2 X1 y1 into Y2 = - X1 s X1 s number X1 X1 sare into y1 [Music] Y and sorry y y both are OPP so y1 POS Y2 postive wrong y y1 negative Y2 positive uh this is correct y1 positive Y2 negative correct y BC D option is only the correct answer BC yes very good [Music] soate of a comple number algebra sum of the complex numbers product of the complex numbers bate of a comp number okay right Z equal x + iy comple number equ X+ iy comple number xus iy for example for example 2us 3 I Z is equal to 3 I + 2 3 I + 2 conjugate z bar 3 I + very good s s skip the class because tomorrow I examine applications of derivatives applications of derivatives okay but this class whatever I'm going to teach you now complex numbers ultimate but today what I'm going to do that concised comple numbers comp uh minus 3 a + 2 2 I minus 3 very good you catched it sir JS exam application application 8 IUS 2 equal to 8 IUS 2al 8 IUS so what is the use of area are question important okay number 22 homework homework okay 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 so right answer very good very good right so right problem comp bar bar conjugate of conjugate is nothing but the complex number comp of the comp number right the complex number this lies in the quadrant number Quant a square minus B Square 1 + 1 2 multiply real into real 1 imaginary into imaginary minus 2 so this is 2 I + I 3 I so what is z this is - 1 + 3 I 2 2 the real number- 1 by imag 3 second second quadrant second very good very good super what is the use of okay second prev complex number comple number conjugates to each other both are conjugates to each other J one and sin X+ I cos 2x is equal to Z2 Bar Z plus I comp imagary imag sin x = cos x and cos 2x = sin 2x 2x 2x possible not possible tan infity infity so not equ so C there is no solution X is a null set okay so there is no such solution very very very important model previous question find out the value of theta for which the expression is real number the number Im it means imagary comp number a 1 I by 1 - 2 I 1 2 IP simple complex numb equal denominator in the a square - b square and a sare 1 + 4 cos sare Theta always into real imaginary into imaginary 2 cos Square Theta imaginary real real imaginary plus 3 I cos Theta so imagary imaginary part of 3 cos th by this 3 cos Theta by I 3 cos thet by this this is what 1 + 4 cos imagary cos Theta thet 2 2 n + 1 into where n is an integer that's it 2 n + 1 by 2 2 n + 2 n + 2 2 odd number into Pi by2 Z any doubt in this real number and imaginary part zero imaginary part cor uh 2 I cos Theta into 1 2 cos Theta cos theta 3 cos Theta right cos Theta 0 Theta by2 32 5 real values of theta that is the correct answer clear right Sim so what is question number two sum of all the possible value of theta belongs to minusa 2 pi for which this is purely imaginary purely imaginary very good purely imaginary IM imal 1 I cos Theta by 1 - 2 I cos Theta the name 1 + 2 I IM question 1 + I thet 1 + I th 1 - 2 cos thet same question 1 + 1 1 plus and 1 minus I into i i sare 2 cos Square thet 3 I 2 cos thet by 1 + 4 cos thet 1 - 2 cos s Theta 0 and cos sare Theta 1 by = 1 by < tk2 S cos by 4 cos s by 4 cosare thet = cos sare Al Theta = n plusus where is integer equations all thees forus right next n = +us 3A n 2 2 2 + sorry 1 + 4 - 4 - 4 3K by 4 5 +k by 4 5K by 4 comma n = 2 pi 2 Pi + 4 2 2us by 4 7 by 4 + 7 Pi by 4 and which is 12 Pi by 4 which is 3 Pi so sum is nothing but how much 3 amay no first first right 3 is the correct answer next question number three if the set real of this and real ofal 3 uh is equal to interval Alpha comma beta then 24 beta minus Alpha is how much and real part of this and real part of real part three real of this is equal to the interval if the set is equal to the interval set Z - z bar + z into z bar by 2 - 3 Z + 5 that is very very very important property mod s that is z into z bar mod Z Square Z into z bar mod and y x² + y s important okay right so Z 2 i y into Z x² real part of zus z bar 2 i y plus Z into Z x² + y² by 2 into 2 2 - 3 Z + 5 z bar uh real of z z and x + I real of Z value 3 zal 3 + I 3 + iy z 3 + I 2 - 3 Z 3 + I + 5 into z 3us i 2 - 9 - 3 i y + 15 - 5 i y so 15 and 6 8 8us 8 I so denominator 8 Comm [Music] + y + 2 i y by 8 * of 1 - i y so real part of 9 + y² + 2 i y 8 * of 1 - iy into 1 + iy MP conjugate denominator na okay important s right soat denominator okay numerator numerator right so real part of numerator real into real 9 + y s IM im- 2 - 2 y by 8 into 1 + y 9 - Y 2 by 8 into 1 + y y s by 8 * of 1 + y s right 10 by 8 1 by 8 10 by 1 + Yus 1 + y 1 + y number y s belongs to Clos Z Infinity perf Square 1 + y songs to closed 1A Infinity reci 1 by 1 + y sare belongs to 0 to closed one 1 + y belongs to open 0a closed 10 10 by 1 + y-s 1X 8 into 10 by 1 + Y 2 - 1 - 1 by 8 plusa 9 by bet Al beta 24 * of beta beta 9 by 8us Alpha minus orus 1 by 8 is + 1 by 8 so 24 into 10 by 8 yes D option is only the correct answer clear very simple very simple option D is the correct answer so question number comp which of the following is Notre IM number Im two of x + i y - 3 I by 4 * of x + i y + 2 y it J so Z is equal to 2x + I into 2 y - 3 by 4X uh + I into 4 y + 2 imag imaginary part 4x - I into 4 y + 2 multip divide IM Im so imaginary of Z so imaginary of 2x into this minus 2x into 4 y + 2 real into imaginary plus imaginary into real 4X into 2 y - 3 0- 2 y + + x into 2 y - 3 0 - 2x y - x + 2xy - 3x is equal to0 2x y 2x y gone 4X = 0 x = = equal i x z x² [Music] 8 x² imagin imaginary plus 2 * of 2 y - 3 into 2 y + 1 so a 8 x² plus 2 * of 4 y s and 8 y s uh - 3 y - 6 y + 2 y - 4 Y and - 8 y - 3 and - 6 4² + 4 Y 2 - 4 y - 3 not equal to 3 and x² + y s - y x² + y² - y not = to 3 y4 x0 y² - y not = 3x4 y not equal 0 y - 1 not = 3x4 I'm sorry y- y 3x4 1 by 4 + 1 by 2us y - 6+ 8 y 4 y 2 y minus 3 6 y plus plus C plus and Y value so y Val 1+ 1 not equal 3 so 3x4 not equal to 3x4 Yus 1 X Y valus which is not correct not correct correct xal 0 correct y + x² + y s 4 not = - 1 by 4 problem Yus IM imary B option should be the not correct answer yes not correct option very good right C question number five question number five right so let the complex number Z equal to x + i y be such that 2 Z - 3 y 3 I by 2 Z + I is purely imaginary if x + y s 0 then what is the value of y^ 4 + Y 2 - 5 okay right one by one Sol just - 3 I by Z 2 * of x + i y + I so Z is equal to 2X + I into 2 y - 3 same question by 2x + 2 y + I 2 y + 1 into I correct 2x + 2 y + I 2 y + 1 2 y - 3 correct so conjugate 2x - I into 2 + into real 4x² imaginary into imaginary plus i into- i - i s and + 1 so this is 2 y - 3 into 2 y + 1 = 0 and the 4 x² + 4 y² - 6 y + 2 y - 4 y - 4 y - 3 = 0 X Y X Y so x² 4 y^ 4 y² minus y = 3 y 4 + y 3 clear conditions fourwheeler father foref fathers sir good news begins in 30 minutes question number3 1 + 2 th 1us imag of the elements in this Z denominator + 2 I I sin Square Theta is equal to sin s by 4 okay and 1 byun sare so Theta equal n + Rus by 4 n pi plus or minus Pi by 4 so from here and Pi byus 3 < - by 4 +k by 4 5 by 4 2 7 by 4 10 16 Pi by 4 and which is 4 Pi 4 Pi is the correct answer similar question that's question number3 a not equal to B to nonzero real numbers then the number of elements in X where real of this is a real of this is b a comp numb so real part x i b sorry real part of a x² - y² + 2 IX y+ BX + i y equal X YX real part of this a - Y 2 + B by a into x = 1 simly y plus a into xal b b x - Y 2 + a by b x = 1 the second this this cancel so a b by a xus a by B xal 0 1 value x = = 0 0 if x = to 0 then what happens - y² is equal to 1 and y sare = to minus1 not possible not possible second B AAL a aareal b a a sare = b square the number of elements in the set xal to so first AAL B AAL B x² y² + x = 1 a = - b x² - y² - x = 180 condition okay right so from here X not possible X Y how to find out y value y value the number of elements in the set a not equal to B condition wrong condition a not equal uh y² = x² - x - 1 b² - 4 A minus - b + orus < TK of b² - 4 a c by 2 a and posi valun 5 - 1 by 2 so y s must be positive y belongs to closed otk 5us 1 by 2 Infinity yare Ys toare root ofun 5 - 1 by 2A Infinity so yely many I question number what is Z power 5 Z into Z 4 square square so what is square square a sare b sare 2 and minus 7 sorry minus 5 - 5 + 2are 25 - 144 minus uh 120 i s 2 + 3 I into 25 45 25 both 20us 119 minus 120 I am I correct am I correct 119 220 120 240 minus 238 imag 360 imaginary into real Justus 300 60 357 357 uh 240 minus 240 so 36 3 38 62 62 and 60 122 minus 6 5 97 I and right 122 plus 597 I + z bar 5 244 so 244 very good 244 is only the correct answer right so right so modulus of a complex number very simp Co an clear right comp mod of squ into Z okay now so mod of sare of no problem we can write like that so 1al 7us 7 + 24 I sare root of 7 Square 49 24 Square 5 76 so 625 625 25 so 25 I mean z comple number very important number is called un sing very important imaginary part of real X imaginary y x y simple properties mod of Z 1 Z2 is mod Z1 into mod Z2 Z 1 by Z2 mod Z 1 by important important mod of Z 1+2 Square Z1 + Z2 into Z1 bar + Z2 bar so modus of Z 1 + Z2 whole square is nothing but mod Z 1 squ + mod Z2 s + 2 * of mod Z 1 into mod Z2 into of theta 1us Theta th Z1 - Z2 into Z1 - Z2 Bar Z 1 barus Z2 bar 1 1+ s s of theta 1us Theta Theta 1us theta0 th 1A th 1 th arent of1 by 0us Maxim that is11 sir 1 + I = 2 49 into x + i y x² y sare equal right simple uh so take mod on both sides mod 1 I 2 I so modus of 1 + I and < tk2 power 100 is equal to 249 square root of x² + y s so < tk2 power 2^ 50 is equal to 2^ 49 root of x² + y sare cancel Ro y X+ y four second option is the correct answer very good next question sir modulus of Z + 2 y less than or equal 2 then greatest value of modulus of Z I modus of minus minus commun3 minus I greatest Max of + 2 I minus root3 233 same + 2 IUS commun3 + I2 the mod of Z1 max1 by that formula less than or equal to mod Z1 Z + 2 I plus mod zun3 + I root 3 + 1 4un 2 maximum fourth option is the correct answer okay now right caran form modulus of Z minus Z KN is equal to some k r is a ccle apprach + 2 I less than oral 200- 2 I cus 2 radius CER 0us 2 0us 2 Center two Radius 0us 2 Center Circle 0us 2 2 units point point modus of zus of < tk3 - I ro33 maximum so p c r what is the maximum distance pc r c Maxim PC Plus R so P30 3 plus -1 plus -1 sare 1 plus r r 2+ 2 that's it comp are so many ways to solve complex numbers complex number question Sol question greatest okay secet less oral mod of + 3 - 5 ofus + 31 and Z + 3 minus 5 modulus of 5 5 so this is always less than or equal to less thanal 2 2 - 5 which isus 3 Maxim uh maximum value sorry is the correct answer third option onec approach cular where the centerus L real number IMX plus r so pance square root of 5 square + 0 square which is 5 so 5 + r r and radius 2 5 option is the correct answer of -4 less than or equal 3 Center -4 0 radius ofus of-1 maximim Maxim pc+ R is the maximum what is the PC three plus r and three which is nothing but 66 answer which is errors best books okay simple 2us 2al 0a 0 P so the least value pcus R mod of pc r root 2 sare + 2 sare 8 minus radius and 1 under root 8 minus 1 fourth option okay right so approach 2 - 2 I = 1 so this is less than or equal to less than or equal to mod Z Plus modus of this under root 8 so mod Z greater oral 1 Maxim max value of this if thisal same question first homework second homework third homework right question number 10 n n question number n p 2024 if R and Theta respect be the modul and amplitude of the complex numbers next 4 + 4 tan S 5 Pi by 8 4 common this 2 * of root of 1 + sare sare 5 secant 5i by 8 5 by [Music] 8 5 Pi by 2 pi minus 3 Pi by uh Pi - 3 Pi by 8 so secant of Pi - 3 Pi second Quantus of 3 by 8 so 2 * of mod ofus of 3 by 8 so modus 2 * of 3 by 8 which is in the first question number 10 2024 question Z be a complex number such that the real part of this is zero then the maximum value of this is equal to how much Max so x + i y - 2 I by x + i y + 2 first of all multiply div the x + I into y - 2 by x + I into y + 2 conjugate x + I into y - 2 into x- I into y + x² + y² - 4 = 0 x² + y² = maximum maximum distance PC Plus R Point P PC Plus R radius so PC 36 + 64 plus r 2 10 + 2 12 12 is the correct answer in the B option geometrical approach geometrical approach sir pyq is how many years 5 years Sol V very good got it hi Amma any doubt Soom what modulus indicates Z1 minus Z2 means what meaning meaning distance between distance between1 and2 meaning ofus Maxim Maximum PC Plus radius R PC Plus r - 6 - 8 x - 6+ I into y - 8 root of x - 6 s + y - 8 12- 16 y plus 6436 perfect way geometrical approach is the perfect way to find answer very easily question number 10 H 10 11 question number 11 J be a complex number such that = cus ofus 2 is equal to 1 Center - 2 comma 0 radius 1 unit then the real Val i y + 1 by x + i y + 2 imaginary part x + 1 + i y by X + 2 + i y x + 2us I this this this x + 2 into y imag - Y into x + imaginary byin x + 2 s + y² equal 1 by 5 imaginary part but XY XY cancel + 2 y - y y so y by x + 2 + y mod of X+ i y + 2 = 1 modulus of x + 2 + i y = 1 = equation x + 2 + y = 1 1al 1 by y value 1X 5 now y Val 1 x + 2 s = 1 - 1 by 25 which is 24 by 25 and x + 2 whole S Plus orus < TK4 and 2un 6 by 5 and x = - 2 + or - 2un 6 by 5 x value + ++ real part X Plus i x + i y + 2 real part x + 2 real part real part x- 2 + Rus 2un 6 by 5 + 2 2 - 2 canc plusus 2un 6 by part mod 2un by answer 2un 6 by 5 a option is the correct answer very good super and hi Amma hi clear 11th question is also done 12 question 2024 Mains s Alpha Beta be the sum and product of the nonzero Solutions of the equation then four * Alpha plus beta square is equal to Alpha sum and product Alpha Su of all the nonzero Solutions of the equation squal Z = - s x + x² + y s minus of z bar and x - i y whole Square so square root of x² + y² = minus CH x² uh plus i s y sare and - y s - 2 IX y so square root of x² + y² is equal to y² - x² + 2 IX y right complex number is equal to complex number - x² square root of x² + y² and 2xy = non zero solution either x0 or y0 x0 or y okay right y y² = to mod Y M y = yal y into y = y yal 1 yal plus orus one - x² = mod x square root of x squ and x so - x Square = x x + x² z x is a real number POS not possible comple numbers IUS betus iuct minus I square that is 1 so Alpha Four * of alpha square + beta square is nothing but four okay now yes very good option A is only the correct answer right question number 12 13 12 13 J 2024 question is a complex number such that less than or equal to one then the minimum value of this is how much cular Center 0 radius one unit minus of 3x2 Z minus of - 3 by2 uh plusus 2 I minus of 1us 3 - 32-2 r 9 + 4 PC minus r and 1 so 16 25 by 4 -1 5 by 2 -1 which is 3x2 3x2 is the correct answer 3 correct answer 14th question modus of zus 1 = 1 this into this minus this = 2 < tk2 Z 1 Z2 are the such that this maximum this is minimum then w 2x zus I right already mod ofus 1al 1 ccle Circle < tk2 -1 into 2x - I into 2 I yal 2 < tk2 2 * of < tk2 - 1 into x - I 2 and + 2 y = 2 < tk2 2 2 2 so < tk2 -1 into x + y = < tk2 i sorry x -1 + i y y what is y < tk2 - < tk2 - 1 into X xun squal so x² - 2x + 1 plus a - b s a s + < tk2 - 1 s b² - 2 a 2 into < tk2 into < tk2 - 1 into x = 1 so x² - 2x + + 3 2 + 1 3 - 2 < tk2 into x² uh minus of minus + 2 < tk2 - 4 into X = 190 x² plus 2un - 6 2 < tk2 - 6 into X x² sorry sorry sorry 3 2un 4 - 2 < tk2 into x² + 2 < tk2 - 6 2 < tk2 - 6 into x + 2 = 0 2 Comm delete 2 - < tk2 x² +un 2 - 3 into x + 1 = = sir < tk2 minus root2 cancel 2 1 3 - 3 x = 1 uh of the roots Al beta c 1 qu qu qu of root c c by 1 by 2 > B < tk2 minus of X into < tk2 minus1 so yun2 > = 1 2 > Y = > 2 minus x value and the 1 by 2 - < tk2 < tk2 - 1 y so y LCM this 2 < tk2 - 2 - < tk2 + 1 by 2 - < tk2 so y = < tk2 - 1 by uh J equal to 1 by 2 - < tk2 + 1 by < tk2 I complex number 1 + 1un + < tk2 by 2 2 + < tk2 by 4 - 2 2 1 + 1 byun2 1 + 1 by 2 + 2 a < tk2 a sare + b square and 1 by 2 < tk2 Comm 1 byun2 minus 1 1 by < tk2 -1 + IUN 1-2 -1 - I modulus of 1 - < tk2 + 1 by < tk2 - 1 s - < tk2 + 1 and 2 - < tk2 2 - > commun2 byun2 square is the correct answer two is the correct answer right root2 Square okay 2k2 2 first the modulus conjugate basic questions of conjugate two comple numbers Z 1 + Z2 is equal to 5 Z 1 CU + Z2 cube is equal to sorry 20 + 15 I then what is the modulus of this z12 a plus b a + b whole Cub minus 3 * of Z 1 Z2 into Z 1 + Z 2 uh 20 + 15 I is equal to 5 CU 125 minus 3 * of Z 1 Z2 into Z 1 + Z2 is equal to 25 f - 3 Z 1 Z and 3 * of Z1 21 3 7us 42 Z1 s + Z2 sare whole Square minus 2 * sare Z 1 + Z2 s - 2 * of Z 1 Z 2 so Z 1 + Z 2 sare 25 minus 2 * of Z 1 Z2 and 14 + uh 2 I 11 + 2 11 + 2 i squ - 2 * of 7 - I Square am I correct right Z 1 power 4 + 2^ 4 121 + - 4 + 44 i - 2 * of 7 - I Square 49 -1 -4 I so 121 is 48 and 90 6 120 117 + 44 I minus 48 96 + 28 I 21 plus 72 I the Z 1^ 4+ 2^ 4 if I don't do any mistake 21 square + 72 square 7222 uh 30 72 11 + 2 11 + 2 44 121 minus 4 44 + 28 72 square 3 * of root of 9 49 + 576 oh three times of 25 I thought what 75 is the correct answer very good very good super cor 75 is the correct answer 15th question okay right mod Square andus I mod mod of = mod ofus I into z i one Z Z okay- mod Z equal 0 mod Z Comm this zus 1 = 0 notal right question number 17 J 2 3- 2ot 9 4 + 4 25 5 by2 25 by 4 and 5 by2 and Val 5 byal Al into 1X 2 - 2 I plus beta + beta I so 5 by 2 equal Alpha by 2 + beta real number beta - 2 into I beta- 2 imag imag imagary beta + Alp by 2 = 5 by 2 Beta - 2 Alp = 0 beta equal 2 Alpha so beta equal to 2 Alpha Beta by 2 isal 2 Bet alal 5 so 2 Beta and 4 Alpha plus alal 5 alal bet sorry three three is only the correct answer simple questions 2024 question number 18 2024 27 January easy shift easy shift so is a complex number such that of zus I is equal to modus of zus ofus I equal to modus of zus 1us 0us one 1A only one yes one is only the correct answer there is only one point like that 18 19 okay very good Z is a complex number z bar is this then modulus of J Square complex numbers x i is I into bar X Plus x - I yal I into s x + i y sare x² - y² + 2 i x y right so x - i y = i * of x² - y sare + x i into I -1 and - 2x Y is real part equal to real part imaginary partal imagus 2X yal and and x² - y² + x = - y so r y y = y y y - = y value comp number second y - y x² 1X 4 + x = 1X 2 and x² + xal 3x 4 so x² + x 3 1 3² + 4 x - 3 = 0 so 4 x² uh plus uh 3x 4X - 3x - 3 = 0 4 4 so so product of the roots alus 3x4 9 by 16us 3x4 Alpha Beta is equal to- 3 1 so bet 3 so 9x 4 - 3x 2 which is 9 - 6X 4 is 3 1 byus 3 okay sorry y by x 1 by 2 - 1X 2 i - 3x2 1 1 by 4 1 by 2 3 9 by 4 + 9 by 4 + 1 by 4 10 by 4 10 by 4 5 by 2 so answer 2 + 6 by 2 so 2 + 3 sorry sorry sorry z z one 1 3 four [Music] option 20 beta mod the square root of alpha + 2 s + beta squ equal Alpha + I beta + 4 + 4 I so real part equal to real part and Al Al + 4 equal Alpha + 4al root of alpha + 2 sare + beta square and beta + 4al 0 and beta value - 4 beta Alp + 4 = square root of uh Alpha + 2 s + 16 squaring on both Alpha sare + 16 + 8 alpha = alpha sare + 4 + 4 Alpha + 16 16 16 gone Alpha Square Alpha Square gone 4 Alpha is = 4 alpha = 1 Alpha is equal 1 1 - 4 and - 3un Alp into beta - 4un -3 -2 Roots qura x² + 7 x + 12 this is the equation D option x² + 7 x + 12 very good super D option is only the correct answer 21 num that the complex number this is unit modulus a plus I lies inside the circle and unit modulus modulus of 1 + a I by B + I should be equal to 1 and 1 + a i modulus equal to of B+ I 1 + a² = b² + 1 a = b² and a equal to plus orus b Bal plus orus aess A and B are of opposite signs opposite sign B valus okay right a I li on the a i b - 1 modulus is equal to 2 * of a + i b modulus a 1 2 + b² = 4 a² + 4 b s balus a a - 1 2 + a² = 4 a² + a square a square - 2 a + 1 is equal to the 7 a sare 6 a + 2 a - 1 = 0 quadratic right so a is = - b + orus Ro TK of b² - 4 a c 24 by 2 a and 12 a = -1 by 6 plus r minus 28 2un 7 by 6un 7 by 6 plus orus two real numbers then the vales of this by 4 Val so 1 + byus 4us uh 1 plus step of AEP for example a value -1 by 6 minus otk 7 by 6 --un 7 by 6-1 2.5 approximately 1.5 2 - 3.5 by 6 3 1us 0 conditions so okay 1 by- 4us 1 by 4us 1 by 4 * of a a and < TK 7 -1 by 6 so 2 2 are 2 3 are - 3 by 2 * of otk 7 - 1 very 21 22 the center and radius of the circle 2al 2 * ofus 3 + iy modulus of x - 2 + i y s is equal to 2 * of modulus of x - 3 + i y s x - 2 2 + Y 2 = 4 * of x - 3² + y s straight l so x² + Y 2 - 4x + 4 it 4 x² + 4 y² - 24x so 3x² + 3 y² - 20x = 36 2 32 circle circle x² + y² - 20 by3 x + 32 by 3us 10 by3 comma Z Alpa beta gamma is the radius GMA radius and R is equal the square root of g s 100 by 9 minus c c 32 by and 33 divide 3 96 by 9 4 2 GMA 2 by3 so 3 * of Al 10 by3 beta 0 GMA 2 by is thect answer yes 12 is only the correct answer question number 22 modulus modulus questions3 the maximum a by zal 2 mod maximum value a sare + 1 + 1 the minimum value square root of a square + 1 Max root2 Maxim minimun 1 by zal 2 of zus 1 by Z is less than or equal + 1 by greater than oral zus 1 by with thisal with thisal modus of -1 by mod Z is less than or equal to 2 mod Z - 1 by Z less than oral 2 great than oral minus 2 s- 2 * of z- 1 less than or equal Z mod z s + 2 * of mod Z - 1 oral zus B+ Rus root of b² - 4 a c by 2 a 1 + Rus 4un belongs to- 1 - > 1un right next argument and amplitude important important comp [Music] number comp number very very important second third important okay now right so okay right what is the argument of a complex number for example complex number that is X Plus i y z complex number r okay but number is so clockwise is same as minus of anticlockwise so minus of anticlockwise sorry that's what anticlockwise is same as minus of clock very simple very simplea is always fromus 5 to 5 first quadrant second quadrant third quadrant fourth quadrant [Music] number second quantal third Quantus fourth qualus Al Alpha alha Al I tan inverse of modulus of Y byx tan inverse y but 30° 30° minus 180 Deus 150 argument of the complex number ude arent both are same right argument of z112 argument of Z 1 byus AR ofus ARG of Z 1 power n argument of z n into Rus 2 important for example number something arent of Theus number number 18us for example examples argument of argument ofus 3 argument of 2 I argument of- 6 I questions argument of positive number number arent of negative number of negative real number Pi argument of positive imaginary Pi by 2 argument of negative imaginary minus 5 so xaxis y AIS complex numb foral 1 first of all I should find out what is Alpha Al tan inverse of modulus of Y by X so Alpha is equal to tan inverse of modulus of Y and I 1 by xus = tan inverse of-1 second so Theta minus Alpha theta 3 by4 so what is the argument of the complex number 3 4 very good super 135 next example tan sorry equal to minus < tk3 uh minus I let me check once answer right3 minus1 a quadrant s third quadrant right first Al call Tan inverse of mod Y byx mod - 1 by - < tk3 alpal tan inverse of 1 byun3 tan inverse of 1 byun3 alha so Theta is nothing but Alpha < by 6- and Theta equal to - 5 pi yes - 1508 2 pi by3 y by X tan inverse of Y by X Y by x 1 byun3 tan 30 30 by 6 Tan 30 and uh 180 30 minus 180 andus 15 very good super very good 1 + < tk3 I into by 1 + I let - 1 + I into uh < tk3 2 < tk3 23 2 argument of Z2 by Z3 so argument of Z argument of Z1 Z2 by Z3 logi argument of numerator minus argument of denominator Al AR Z1 plus AR z23 log X1 X Y log X log y so argument of Z1 plus AR Z2 US arus 1 byun3 1 byun3 1 by root3 root3 by 6 Min - 5 by 6 minus AR Z3 quadrant to third quadrant uh second quadrant second Quantus by 4 3 by 4 answer < by 2 - 3 < by 4 which is - < by4 is the final answer am I correct 60 1us 30 30 Quant second quadrant negative comma positive second quadrant second quadrant the answer 45 but Pi minus 45 and 3 Pi by 4 Pi - by 4 so minus- 3 90 - 3 Pi by 4 which is minus < by 4 very good super I form = R into Theta zal mod Z into cos Theta + I 22 23 J 2023 iusus of 1 - I by cos 60 very sorry cos 60 1 by 2 + I into sin 60 < tk3 by 2 so minus of 1 - I by 1 by 2 2 1 + < tk3 I numerator mod Z 1 by denominator mod Z2 already cos theta plus I Theta so AR Z numerator AR Z1 minus AR Z2 numerator by denominator numerator minus denominator so numerator -1 + I a quadrant netive comma POS second quadrant - by 4 3 4 3i by 4 minus denominator 3 denominator by so argument 12 9 Pius 4 Pi which is 5 Pi 5 Pi by 12a so < tk2 5 by option the argument of so yes D option is only the correct answer right let us move further zal mod Z into e i Theta e for example for example 1un e i this is called form 1K I okay right comp number deleted so next Monday Monday morning okay so that's all for now so special session infity 2. just within 10 minutes 10 to 15 minutes series okay 20 days 80 Days 10 days 60 plus 232 ma class thank you very much math physics chemistry biology telug social physics maybe Monday or Tuesday Monday or Tuesday okay right so that's all for now bye everyone please like the video share the video with friends and please subscribe