InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If `omega(!=1)`is a cube root of unity,and `(1+omega)^7= A + B omega`. Then (A, B) equalsA. `(-1,1)`B. `(0,1)`C. `(1,1)`D. `(1,0)` |
|
Answer» Correct Answer - C `(1+ omega)^(7) = A + Bomega ` ` therefore (-omega^(2))^(7) = A + Bomega` `therefore -omega^(2) = A Bomega ` `therefore 1+ omega = A + Bomega ` `therefore A = 1 , B = 1` ` therefore (A,B) = (1,1)` |
|
| 52. |
If `z!=1`and `(z^2)/(z-1)`is real, then the pointrepresented by the complex number z lies(1)either on thereal axis or on a circle passing through the origin(2)on a circle withcentre at the origin(3)either on thereal axis or on a circle not passing through the origin(4)on the imaginaryaxisA. either on the real axis or on a circle passing thorugh the origin.B. on a circle with centre at the origin.C. either on the real axis or an a circle not possing through the origin .D. on the imaginary axis . |
|
Answer» Correct Answer - A `(z^(2))/(z-1) ` is purely real lt `therefore (z^(2))/(z-1) = (barz^(2))/(barz-1)` ` rArr zbarzz - z^(2) = zbarzbar - barz^(-2)` `rArr (z-barz) (|z|^(2) -( z+ barz))= 0` Either `z = barz` `rArr ` z lies on real axis. or `|z|^(2) = z+z` `rArr zbarz - z - barz = 0` `rArr x^(2) + y^(2) -2x = 0` Which represents a cricle passing throught origin. |
|
| 53. |
If z is a complexnumber of unit modulus and argument q, then `a r g((1+z)/(1+ bar z))`equal(1) `pi/2-theta`(2) `theta`(3) `pi-theta`(4) `-theta`A. `-theta`B. `(pi)/(2 ) - theta`C. `theta`D. `pi-theta` |
|
Answer» Correct Answer - C Since `|z| = 1, z.barz= 1` `therefore arg((1+z)/(1+z)) = arg((1+z)/(1+(1)/(z)))= arg=((1+z)/(z+1))` `=argz = theta` |
|
| 54. |
If `z` is a complex number such that `|z|>=2` then the minimum value of `|z+1/2|` isA. is equal to `(5)/(2)`B. lies in the interval (1,2)C. is strictly gerater than `(5)/(2)`D. is strictly greater than `(3)/(2)` but less than `(5)/(2)` |
|
Answer» Correct Answer - B `|z| ge2` `therefore |z+(1)/(2)||ge|-|(1)/(2)||ge |2-(1)/(2)|ge(3)/(2)` Hence, minimun distance between z and `(-(1)/(2),0)` is `(3)/(2)`. |
|
| 55. |
The equation `|z-i|=|z-1|,i= sqrt(-1)`representsA. a circule of radius `1/2`B. the line passing through the origin with slope 1C. a cicle of radius 1D. the line passing the origin with slope-1 |
|
Answer» Correct Answer - B Let the complex nuber z = x+ iy Also given |z-i|=|z-1| `rArr | x+iy -I |=|x+ iy -1|` `rArr sqrt(x^2+(y-1)^2=sqrt((x-1)^2+y^2)` `[therefore|z| = sqrt((Re(z))^2+Im(z)^2)]` On squaring both sides , we get `x^2+y^2-2y+1=x^2+y^22x+1` `rArr y= x` which of the represents a line through the origin with slope 1. |
|
| 56. |
If `z_1,z_2 and z_3` are complex numbers such that `|z_1|=|z_2|=|z_3|= |1/z_1+1/z_2+1/z_3|=1, then |z_1+z_2+z_3|` is (A) equal to 1 (B) gt1 (C) gt3 (D) equal to 3A. equal to 1B. less than 1C. greater than 3D. equal to 3 |
|
Answer» Correct Answer - A Given `" "|z_1|=|z_2|=|z_3|=1` Now `" "|z_1|=1` `rArr " " |z_1|^2=1 rArr z_1 bar z_1=1` similarly `z_2barz_2=1,z_3 bar z_3=1` Again Now `" "|1/z_1+1/z_2+1/z_3|=1` `rArr |barz_1+barx_2+bar3 |=1rArr |z_1+z_2+z_3=1` `rArr " "|z_1+z_2+z_3|=1` |
|
| 57. |
Acomplex number z is said to be unimodular if . Suppose `z_1`and `z_2`are complex numbers such that `(z_1-2z_2)/(2-z_1z_2)`is unimodular and `z_2`is not unimodular. Then the point `z_1`lieson a :(1)straight line parallel to x-axis(2) straight line parallel to y-axis(3)circle of radius 2(4) circle of radius `sqrt(2)`A. straight line parallel to x-axisB. straight line parallel to y-axiesC. circle of radius 2D. circle of radius `sqrt(2)` |
|
Answer» Correct Answer - C If is unimodular then |z|=1 also use property of modules I,e `zz=|z|^2` Given `z_2` is not unimodular I,e `|z_2|ne 1` and `(z_1-2z_2)/(2-z_1barz_2)|` is unimodular `rArr |(z_1-2z_2)/(2-z_1barz_2)|=1 rArr |z_1-2z_2|^2=|2-z_1barz_2|^2` `rArr ( z_1-2z_2)(barz_1-2barz_2)=(2-z_1bar z_2)(2-barz_1z_2)" "[zbarz=|z|^2]` `rArr |z|^2+4|^2-2barz_1z_2-2z_1bar z_2` `rArr 4+|z_1|^2|z_2|^2-2barz_1z_2-2z_1barz_2rArr(|z_2 |^2-1)(|z_1|^2-4)=0` `because " "|z_2|ne1` `therefore |z_1|=2` Let `z_1=x+iy rArr x^2+y^2=(2)^2` `therefore`Point `z_1`lies on circle of radius |
|
| 58. |
If `a gt 0 and z=(1+i)^2/(a- i)` has magnitude `sqrt(2/5) then barz is ` equal toA. `1/5-3/5 i`B. `-1/5-3/5 I `C. `1/5+3/5 I `D. `-3/5-1/5 I ` |
|
Answer» Correct Answer - B The given complex number `z=((1+i)^2)/(a-i)` `=((1-1+2i)(a+i))/(a^2+1)" "[ therefore i^2 = -1]` `=(2i(a+i))/(a^2+1)=(-2+2 ai )/(a^2+1)` `therefore | z|= sqrt(2//5)` `rArr sqrt(4+4a^2)/(a^2 +1)^2=sqrt(2/5)rArr (2)/(sqrt(1+a^2)=sqrt(2/5)` `rArr (4)/(1+a^2)= 2/5 rArr a^2+1 = 10 ` `rArr a^2=9 rArr a=3` `therefore z=(-2 + 6 i)/(10) " " [From Eq.(i)]` So, `barz=((-2+6i)/(10))=(-1/5+3/5i)rArr barz=-1/5-3/5i` `[ therefore if z= x+ iy , then bar z x-iy]` |
|
| 59. |
If `k>0`, `|z|=w=k`, and `alpha=(z-bar w)/(k^2+zbar(w))`, then `Re(alpha)` (A) 0 (B) `k/2` (C) `k` (D) None of these |
|
Answer» Correct Answer - A `alpha=(z-barw)/(k^(2)+zbarw)rArr baralpha=(barz-w)/(k^(2)+barzw)` But `zbarz =wbarw =k^(2)`. Hence, `rArr " "baralpha=((k^(2))/(z)-(k^(2))/(barw))/(k^(2)+(k^(2))/(z)(k^(2))/(barw))=(barw-z)/(zbarw+k^(2))=-alpha` `rArralpha+baralpha=0` `rArr Re(alpha)=0` |
|
| 60. |
Let z be a complex number such that |z| + z = 3 + i (Where `i=sqrt(-1))` Then ,|z| is equal toA. `sqrt(34)/3`B. `5/3`C. `sqrt(41)/4`D. `5/4` |
|
Answer» Correct Answer - B We have `|z|+z=+i` Let z=x+iy `therefore sqrt(x^2+y^2)+x+iy+3+i` `rArr (x+sqrt(x^2+y^2))+iy=3+i` `rArr x+sqrt(x^2+y^2)=3 and y=1` Now , `sqrt(x^2+1)=3-x` `rArr x^2+1=9-6x+x^2` `rArr 6x=8 rArr x=4/3` `therefore z=4/3+i` `rArr |z|=sqrt(16/9)+1=sqrt(25/9) rArr |z|=5/3` |
|
| 61. |
Let `z_1 and z_2` be two complex numbers satisfying `|z_1|=9` and `|z_2-3-4i|=4` Then the minimum value of `|z_1-Z_2|` isA. 1B. 2C. `sqrt(2)`D. 0 |
|
Answer» Correct Answer - D Clearly `|z_1|=9 ` represents a circle having centre `C_1(0,0)` and radius `r_1=9` and `|z_2-3-4i|=4` represents a circle having centre `C_2(3,4)` and radius `r_2=4` The minimum value of f`|z_1-z_2| ` is equals to minimum distance between circless `|z_1|=9` and `|z_2-3-4i|=4` `therefore C_1 C_2= sqrt((3-0)^2(4-0)^2)= sqrt(25)=5` and `|r_1-r_2|=|9-4|=5 rArr C_1C_2 = |r_1-r_2|` `therefore `Circles touches each other internally . Hence , `|z_1-z_2|_(min)=0` |
|
| 62. |
If `|z| =1 ` and `w=(z-1)/(z+1)` (where `z != -1`) then `Re(w)` is (A) 0 (B) `-1/|z+1|^2` (C) `|z/(z+1)| 1/|z+1|^2` (D) `sqrt2/|z+1|^2` |
|
Answer» Correct Answer - A Since `|z| =1 and w = (z-1)/(z+1)` `rArr z-1 =wz+w rArr =(1+w)/(1-w)rArr |z|=(|1+w|)/(|1-w|)` `rArr |1-w|=|1+w| " "[ becaue |z|=1]` ltbr. On squaring both sides we get `1+ |w|^2-2|w| Re (w) =1 + |w|^2+2|w|Re (W)` [using `|z_1pm z_2|^2=|z_1|^2+|z_2|^2pm 2 |z_1||z_2| Re (bar z_1 z_2)]` `rArr 4 |w|Re |w|=0` `rArr " " Re (w)=0` |
|
| 63. |
Let `(z-alpha)/(z+alpha)` is purely imaginary and `|z|=2, alphaepsilonR` then `alpha` is equal to (A) `2` (B) `1` (C) `sqrt2` (D) `sqrt3`A. `sqrt(2)`B. `1/2`C. 1D. 2 |
|
Answer» Correct Answer - D Since the complex number `(z-a)/(z+a)(alpha ne R)`j is purely imaginary number therefore `(z-alpha)/(z+alpha)+(barz-alpha)/(bar z + alpha)=0 " "[ therefore alpha in R]` `rArr zbarz-abarz+az-alpha^2+zbarz-alphabarz-alpha^2=0` `rArr 2|z|^2-2alpha^2=0" "[because zbarz=|z|^2]` `rArr alpha^2=|z|^2=4` `rArr alpha=pm2` |
|
| 64. |
The expression `[(1+sin(pi/8)+icos(pi/8))/(1+sin(pi/8)-icos(pi/8))]^8` is equal isA. 1B. `-1`C. iD. `-i` |
|
Answer» Correct Answer - B Let `sin""(pi)/(8)+icos""(pi)/(8)=z` `rArr [(1+sin""(pi)/(8)+icos""(pi)/(8))/(1+sin""(pi)/(8)-icos""(pi)/(8))]^(8)` `=((1+z)/(1+(1)/(z)))^(8)=z^(8)` `=(sin""(pi)/(8)+icos""(pi)/(8))^(8)` `=(cos""((pi)/(2)-(pi)/8)+isin""((pi)/(2)-(pi)/(8)))^(8)` `=(cos""(3pi)/(8)+isin""(3pi)/(8))^(8)=cos3pi=-1` |
|
| 65. |
Given `z=(1+isqrt(3))^(100),`then `[R E(z)//I M(z)]`equals`2^(100)`b. `2^(50)`c. `1/(sqrt(3))`d. `sqrt(3)`A. `2^(100)`B. `2^(50)`C. `(1)/(sqrt(3))`D. `sqrt(3)` |
|
Answer» Correct Answer - C `z=(1+isqrt(3))^(100)=2^(100)(cos""(pi)/(3)+isin""(pi)/(3))^(100)` `=2^(100)(cos""(100pi)/(3)+isin""(100pi)/(3))` `=2^(100)(-cos""(pi)/(3)-isin""(pi)/(3))` `=2^(100)(-(1)/(2)-(isqrt(3))/(2))` `rArr (Re(z))/(Im(z))=(-1//2)/(-sqrt(3)//2)=(1)/(sqrt(3))` |
|
| 66. |
Let z be a complex number such that the imaginary part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value(A) –1 (B)1 3(C)1 2(D)3 4A. `-1`B. `1/3`C. `1/2`D. `3/4` |
|
Answer» Correct Answer - D IF `ax^2+bx+c=0` has roots `alpha ,beta ` then `alpha,beta=(-bpmsqrt(b^2-4ac))/(2a)` For roots to be real `b^2-4ac ge0` ltbgt Description of situation As imeginary part of z=x+iy is non-zero `rArr y ne 0 ` Method I let z=x+iy `therefore " "a=(x+iy)^2+(x+iy)+1` `rArr (x^2-y^2+x+1-a)+i(2xy+y)=0` `rArr (x^2y^2+x+1-a)+iy(2x+1)=0` If is purely real if y(2x+1)=0 but imaginary part of z.ie y is non-zero `rArr 2x+1=0 or x=-1//2` From Eq.(i) `1/4-y^2=1/2+1a=0` `rArr a=-y^2+3/4 rArr a lt 3/4` Method II `Here z^2+z+(1-a)=0` `therefore " "z=(-1pmsqrt(1-4(1-a)))/(2xx1)` `rArr " "z=(-1pmsqrt(4a-3))/(2)` For z do not have real roots `4a-3 lt 0 rArr a lt 3/4` |
|
| 67. |
Find the common roots of `x^12-1=0 and x^4+x^2+1=0` |
|
Answer» Correct Answer - `x + pm omega^(2), pm omega` `x^(1//2)-1=(x^(6) + 1)(x^(6)-1)=(x^(6)+1)(x^(2)-1)(x^(4)+x^(2)+1)` Common roots are given by `x^(4) + x^(2) + 1=0` `therefore " " x^(2)=(-1pm,isqrt(3))/(2) = omega,omega^(2) or omega^(4),omega^(2)" "(because omega^(3)=1)` ` or x = pm omega^(2), pm omega` |
|
| 68. |
If `z + z^(-1)= 1`, then find the value of `z^(100) + z^(-100)`. |
|
Answer» Correct Answer - `-1` `z +z ^(-1)=1` ` or z^(2) -z + 1=0` `rArr z =- omega or -omega^(2)` For `z = -omega` `z^(100) + z^(-100) = (-omega)^(100) + (-omega)^(100)` `= omega+(1)/(omega) = omega + omega^(2)=1` For `z = -omega^(2)`, `z^(100) + z^(-100)=- (-omega^(2))^(100) _ (-omega^(2))^(-100)` ` = omega^(200) + (1)/(omega^(200))` `= omega^(2)+(1)/(omega^(2)) = omega^(2) + omega = -1` |
|
| 69. |
Write the comple number in a + ib form unsing cube roots of unity: (a) `(-(1)/(2) + sqrt(3)/(2)i)^(1000)`(b)If `z =(sqrt(3) + i)^(17)/((1-i)^(50))` (c) `(i + sqrt(3))^(100) + (i+ sqrt(3))^(100) + 2^(100)` |
|
Answer» Correct Answer - `omega = - (1)/(2) + sqrt(3)/(2)i`" " (b) `(1)/(2^(8))((-1-isqrt(3))/(2))` " "(c) 0 (a) Here, `-1//2 + (1//2) isqrt(3)` is one of the tow imaginary cube roots of unity. If the we denote it b `omega`, then `omega^(100) = omega^(999) omega=(omega^(3))^(333) omega = omega = -(1)/(2) + (sqrt(3))/(2) i` `z =(sqrt(3) + i)^(17)/((1-i)^(50))` `= (1)/(i^(17))((isqrt(3) +i^(2))^(17))/([(1-i)^(2) ]^(25))` `= (2^(17))/(i) (((isqrt(3)-1)/(2))^(7))/(-2i)^(25)` `=(1)/(2^(8)) (omega)^(17)` `= (1)/(2^(8)) (omega)^(2)` `= (1)/(2^(8)) ((-1-isqrt(3))/(2))` (c) `(i+sqrt(3))^(100) + (i-sqrt(3))^(100) + 2^(100)` `=((i^(2) + isqrt(3))/(i))^(100) + ((i^(2)-isqrt(3))/(i))^(100) + 2^(100)` `= (2^(100))/(i^(100)) ((-1+isqrt(3))/(2))^(100) + (2^(100))/(i^(100)) ((-1-isqrt(3))/(2)) +2^(100)` `= 2^(100) (omega) ^(100) + 2^(100) (omega^(2))^(100) + 2^(100)` `=2^(100) (omega^(100) + omega^(200)+1)` ` = 2^(100) (omega + omega^(2) + 1) =0` |
|
| 70. |
Let a,b ` in ` R and `a^(2) + b^(2) ne 0` . Suppose `S = { z in C: z = (1)/(a+ ibt),t in R, t ne 0}`, where `i= sqrt(-1)`. If `z = x + iy` and z in S, then (x,y) lies onA. the circle with radius `(1)/(2a)` and centre `((1)/(2a),0)` for `a gt 0 be ne 0`B. the circle with radius `-(1)/(2a)` and centre `(-(1)/(2) ,0) a lt 0, b ne 0`C. the axis for `a ne 0, b =0`D. the y-axis for `a = 0, bne 0` |
|
Answer» Correct Answer - A::C::D `z =(1)/(a+ibt)` `rArr x +iy = (a-ibt)/(a^(2) + b^(2)t^(2))` `rArr x = (a)/(a^(2) + b^(2)t^(2)),y = (-bt)/(a^(2) =b^(2)t^(2))` Eliminating t, we get `x^(20 + y^(2) = (x)/(a)` `rArr (x-(1)/(2a))^(2) +y^(2) = ((1)/(2a))^(2)` `therefore `Option (1) is correct. (3),(4) can be verified by putting b = 0 and a=0 respectively. |
|
| 71. |
Which of the following is equal to `root(3)(-1)?`A. `(sqrt(3)+sqrt(-1))/( 2)`B. `(-sqrt(3)+sqrt(-1))/(sqrt(-4))`C. `(sqrt(3)-sqrt(-1))/(sqrt(-4))`D. `-sqrt(-1)` |
|
Answer» Correct Answer - B `x =root(3)(-1)` `rArr x^(3)=-1` `rArr -x^(3)=1` `rArr-x=1,omega,omega^(2)` ltbgt `rArr x=-1,-omega,-omega^(2)` `=-1,(1+sqrt3i)/(2),(1-sqrt3i)/(2)` `=-1,(-sqrt3+i)/(2i),(sqrt3+i)/(2i)` `=-1,(-sqrt3+sqrt(-1))/(sqrt(-4)),(sqrt(3)+(sqrt(-1)))/(sqrt(-4))` |
|
| 72. |
Let P(x) and Q(x) be two polynomials.Suppose that `f(x) = P(x^3) + x Q(x^3)` is divisible by `x^2 + x+1,` thenA. P(x) is divisible by (x-1),but Q(x) is not divisible by x -1B. Q(x) is divisible by (x-1), but P(x) is not divisible by x-1C. Both P(x) and Q(x) are divisible by x-1D. f(x) is divisible by x-1 |
|
Answer» Correct Answer - C::D We have `x^(2) + x+1 = (x-omega)(x- omega^(2))` Simce `f(x)` is divisible by `x^(2) + x + 1,f(omega)=0, f(omega^(2)) = 0` so `p(omega^(3)) + omegaQ(omega^(3)) = 0 rArr p(1) + omegaQ(1) = 0" "(2)` `P (omega^(6)) + omega^(2)Q(omega^(6)) = 0 rArr P(1) + omega^(2) Q(1)= 0 " (2)` Solving Eqs. (1) and (2), we obtain `P(1) = 0 and Q(1) = 0` Therefore, both P(x) and Q(x) are divisble by x-1. Hence, `P(x^(3)) and Q(x^(3))` are divisible by `x^(3) -1` and soby x -1 Since `f(x) = P(x^(3)) + xQ(X^(3))`, we get f(x) is divisible by x-1 |
|
| 73. |
If `x^2+x+1=0` then the value of `(x+1/x)^2+(x^2+1/(x^2))^2+...+(x^27+1/(x^27))^2` isA. 27B. 72C. 45D. 54 |
|
Answer» Correct Answer - D `x^(2)+x+1=0` `rArr x = omegaor omega^(2)` Let `x = omega`. Then, `x+(1)/(x)=omega+(1)/(omega)=omega+omega^(2)=-1` `x^(2)+(1)/(x^(2))=omega^(2)+(1)/(omega^(3))=omega^(2)+omega=-1` `x^(3)+(1)/(x^(3))=omega^(3)+(1)/(omega^(3))=2` `x^(4)+(1)/(x^(4))=omega^(4)+(1)/(omega^(4))=omega+omega^(2)=-1,` etc. `therefore (x+(1)/(x))^(2)+(x^(2)+(1)/(x^(2)))^(2)+(x^(3)+(1)/(x^(3)))^(2)+* * *+(x^(27)+(1)/(x^(27)))^(2)` `=[(x+(1)/(x))^(2)+(x^(2)+(1)/(x^(2)))^(2)+(x^(4)+(1)/(x^(4)))^(2)+* * *+(x^(26)+(1)/(x^(26)))^(2)]` `+[(x^(3)+(1)/(x^(3)))^(2)+(x^(6)+(1)/(x^(6)))^(2)+(x^(9)+(1)/(x^(9)))^(2)+* * *+(x^(27)+(1)/(x^(27)))^(2)]` `=18+9(2)^(2)=54` |
|
| 74. |
If `5x^(3) +Mx+ N, M,N in R` is divisible by `x^(2) + x+1`, then the value of `M +N ` isA. 5B. 4C. `-4`D. `-5` |
|
Answer» Correct Answer - D Let `f(x)=5x^(2)+Mx+N`. Also, `x^(2)+x+1=(x-omega)(x-omega^(2))` It is given that `f(x)` is divisible by `x^(2)+x+1`. `therefore f(omega)=5+Momega+N=0` and `f(omega^(2))=5+Momega^(2)+N=0` So, `M =0 and N=-5` `therefore M+N =-5` |
|
| 75. |
Sum of common roots of the equations `z^(3) + 2z^(2) + 2z + 1 =0` and `z^(1985) + z^(100) + 1=0` isA. `-1`B. 1C. 0D. 1 |
|
Answer» Correct Answer - A We have, `z^(3)+2z^(2)+2z+1=0` `rArr (z^(3)+1)+2z(z+1)=0` `rArr (z+1)(z^(2)+z+1)=0` `rArrz=-1,omega,omega^(2)` Since, `z=-1` does not satisfy `z^(1985)+z^(100)+1=0` while `z=omega, omega^(2)` satisfy it, hence, sum is `omega+omega^(2)=-` |
|
| 76. |
Let a be a complex number such that `|a| lt 1` and `z_(1),z_(2)…..` be vertices of a polygon such that `z_(k)=1+a+a^(2)+a^(3)+a^(k-1)`. Then, the vertices of the polygon lie within a circle.A. `|z-(1)/(1-a)|=(1)/(|a-1|)`B. `|z+(1)/(a+1)| = (1)/(|a+1|)`C. `|z-(1)/(1-a)|=|a-1|`D. `|z+(1)/(1-a)|=|a-1|` |
|
Answer» Correct Answer - A Given, `z_(k) = 1 + a+ a^(2) +......+a^(k-1) = (1-a^(k))/(1-a)` `rArr a_(k) -(1)/(1-a) = - (a^(k))/(1-a)` `rArr |z_(k)-(1)/(1-a)|= (|a|^(k))/(|1-a|) lt (1)/(|1-a|)" " [because|a| lt 1]` Hence , `z_(k)` lies within the circle. `therefore |z-(1)/(1-a)|= (1)/(|1-a|)` |
|
| 77. |
If `n in N >1`, then the sum of real part of roots of`z^n=(z+1)^n`is equal to`n/2`b. `((n-1))/2`c. ` n/2`d. `((1-n))/2`A. `(n)/(2)`B. `((n-1))/(2)`C. `-(n)/(2)`D. `((1-n))/(2)` |
|
Answer» Correct Answer - D The equations `z^(n) = (z+ 1)^(n)` will have excactly n - 1 roots. We have `((z+1)/(z^(n)))= 1 or |(z+1)/(z)|= 1` or `|z+1|= |z|` Therefore,z lies on the the right bisector of the segment connecting the points (0,0) and (-1,0). Thus `Re(z) = - 1//2`. Hence roots are colliner and will have their real parts of roots is `(-1//2)(n-1)` |
|
| 78. |
If `z(1+a)=b+i ca n da^2+b^2+c^2=1,`then `[(1+i z)//(1-i z)=``(a+i b)/(1+c)`b. `(b-i c)/(1+a)`c. `(a+i c)/(1+b)`d. none of theseA. `(a+ib)/(1+c)`B. `(b-ic)/(1+a)`C. `(a+ic)/(1+b)`D. none of these |
|
Answer» Correct Answer - A `(1+iz)/(1-izx)=(1+i(b+ic)//(1+a))/(1-i(b+ic)//(1+a))` `=(1+a-c+ib)/(1+a+c-ib)` `=((1+a-c+ib)(1+a+c+ib))/((1+a+c)^(2)+b^(2))` `=(1+2a+a^(2)-b^(2)-c^(2)+2ib+2iab)/(1+a^(2)+c^(2)+b^(2)+2ac+2(a+c))` `=(2a+2a^(2)+2ib+2iab)/(2+2ac+2(a+c))" "(therefore a^(2)+b^(2)+c^(2)=1)` `=(a^(2)+a^(2)+ib+iab)/(1+ac+(a+c))` `=(a(a+1)+ib(a+1))/((a+1)(c+1))=(a+ib)/(c+1)` |
|
| 79. |
If `a^2+b^2=1` then `(1+b+i a)/(1+b-i a)=`A. 1B. 2C. 5D. 8 |
|
Answer» Correct Answer - C Given that `a^(2) +b^(2)=1` Therefore, `(1+b + ia)/(1+b -ia) =((1+b+ia)(a+b+ia))/((1+ b-ia)(1+b+ia))` `= ((1+b)^(2) -a^(2)+2ia(1+b))/(1+ b^(2)+2b+a^(2))` `= ((1-a^(2)) + 2b b^(2) + 2ia(1+b))/(2(l+b))` `= (2b^(2) + 2b + 2ia(1+b))/(2(1+b))= b + ia` |
|
| 80. |
If `alpha`and `beta`are different complex numbers with `|beta|=1,`then find `|(beta-alpha)/(1- baralphabeta)|`. |
|
Answer» Given, `|beta|=1` `implies" "betabar(beta)=1" or "beta=(1)/(beta)" "(1)` Now, `|(beta-alpha)/(1-bar(alpha)beta)|=|((1)/(bar(beta))-alpha)/(1-bar(alpha)beta)|=|(1)/(bar(beta))||(1-alphabar(beta))/(1-bar(alpha)beta)|=(1)/(|beta|)|bar(1-alphabar(beta))/(1-bar(alpha)beta)|` `=|(1-bar(alpha)beta)/(1-bar(alpha)beta)|=1` |
|
| 81. |
Solve the equation `z^(3) = barz (z ne 0)` |
|
Answer» As `zne0,` `|z^(3)|=|bar(z)|` `implies" "|z|^(3)=|bar(z)|` `implies" "|z|^(3)=|z|` `implies" "|z|=1` `:." "z^(3)=bar(z)` `implies" "z^(4)=zbar(z)=|z|^(2)=1` `implies" "z=+-1,+-i` |
|
| 82. |
If `z_1, z_2 in C , z_1^2 in R , z_1(z_1^2-3z_2^2)=2` and `z_2(3z_1^2-z_2^2)=11`, then the value of `z_1^2+z_2^2`isA. 10B. 12C. 5D. 8 |
|
Answer» Correct Answer - C We have, `z_(1)(z_(1)^(2)-3z_(2)^2) = 2 " "(1)` `z_(2)(3z_(1)^(2)-z_(2)^(2)) = 11` multiplying (2) by I and adding it to (1), we get `z_(1)^(3) - 3z_(2)^(2)z_(1) + i(3z_(1)^(2)z_(2) - z_(2)^(3)) = 2 +11i` `rArr (z_(1) +iz_(2))^(3) = 2 + 11i" "(3)` Multiply (2) by i and subtracting it form (1), we get `z_(1)^(3) - 3z_(2)^(2)z_(1) - i(3z_(1)^(2)z_(2)-z_(2)^(2)) = 2- 11i` `rArr (z_(1) + iz_(2))^(3) = 2+ 11i` Multiplying (3) and (4), we get `(z_(1)^(2) + z_(2)^(2))^(3) = 2^(2) - 121i^(2) = 4 + 121 = 125` `rArr z_(1)^(2) + _(2)^(2) = 5` |
|
| 83. |
`Z_1!=Z_2`are two points in an Argand plane. If `a|Z_1|=b|Z_2|,`then prove that `(a Z_1-b Z_2)/(a Z_1+b Z_2)`is purely imaginary. |
|
Answer» Let `Z_(1) = r_(1) e^(itheta),Z_(2) =r_(2)e^(i(theta + alpha))` Given that `ar_(1) = br_(2)` `therefore Z = (aZ_(1) - bZ_(2))/(aZ_(1) + bZ_(2))=(e^(itheta)-e^(i(theta + alpha)))/(e^(itheta) + e^(itheta+alpha))` `= (1-e^(ialpha))/(1+e^(ialpha))" "("Dividing Nr. and Dr. by" e^(itheta))` `(e^(-ialpha//2)-e^(ialpha//2))/(e^(-ialpha//2) + e^(ialpha//2))" "("Dividing Nr. and Dr.by " e^(ialpha//2))` `= (-2i sin.(alpha)/(2))/(2cos.(alpha)/(2))` `=-i tan .(alpha)/(2)` Hence, Z is purely imaginary. |
|
| 84. |
Prove that `|z_1+z_2|^2=|z_1|^2, ifz_1//z_2`is purely imaginary. |
|
Answer» Given `|z_(1)+z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)` or `|z_(1)|^(2)+|z_(2)|^(2)+z_(1)bar(z)_(2)+bar(z)_(1)z_(2)=|z_(1)|^(2)+|z_(2)|^(2)` or `z_(1)bar(z)_(2)+bar(z)_(1)z_(2)=0` or `(z_(1))/(z_(2))+(bar(z)_(1))/(bar(z)_(2))=0` or `(z_(1))/(z_(2))+(bar(z)_(1))/(z_(2))=0` Hence, `(z_(1))/(z_(2))` is purely imaginary |
|
| 85. |
Let complex numbers `alpha and 1/alpha` lies on circle `(x-x_0)^2(y-y_0)^2=r^2 and (x-x_0)^2+(y-y_0)^2=4r^2` respectively. If `z_0=x_0+iy_0` satisfies the equation `2|z_0|^2=r^2+2` then `|alpha|` is equal to (a) `1/sqrt2` (b) `1/2` (c) `1/sqrt7` (d) `1/3`A. `1//sqrt(2)`B. `1//2`C. `1//sqrt(7)`D. `1//3` |
|
Answer» Correct Answer - C Given circles are `(x - x_(0))^(2) + (y-y_(0))^(2) = r^(2)` and `(x-x_(0))^(2) + (y - y_(0))^(2) = 4r^(2)" "(1)` or `|z-z_(0)| = r` and `|z-z_(0)| = 2 r` where `z_(0) - x_(0) + iy_(0)` Now `alpha` and `(1)/(baralpha)` lies on circle (1) and (2), respectively. Then ` |alpha -z_(0)| = r and |(1)/(baralpha) - z_(0)| = 2r` `rArr |alpha - z_(0)| = r and |1-baralphaz_(0)| = 2r |baralpha|` `rArr |alpha -z_(0)| = r and |1-baraz_(0)| = 2r|baralpha|` `rArr |alpha - z_(0)|^(2) =r^(2) and |1-baralphaz_(0)| = 4r^(2) |alpha|^(2)` Subtracting, we get `|1-baralphaz_(0)|^(2) - |alpha -z_(0)|^(2) = 4r^(2) |alpha| - r^(2)` `rArr 1+ |alphaz_(0)|^(2) -baralphaz_(0) - alphabarz_(0) -(|alpha|^(2) + |z_(0)|^(2) -baralphaz_(0) -alphabarz_(0))` `4r^(2) |alpha|^(2) -r^(2)` `rArr 1+|alpha|^(2) |z_(0)|^(2) - |alpha|^(2) -|z_(0)|^(2) = 4r^(2)|alpha|^(2)-r^(2)` Given ` 2|z_(0)|^(2) = r^(2) + 2` `2|z_(0)|^(2) = r^(2)+ 2` `rArr (1-|alpha|^(2))(1-(r^(2) +2)/(2)) = 4r^(2)|alpha|^(2) -r^(2)` `rArr (1-|alpha|^(2))((-r^(2))/(2))= 4r^(2)|alpha|^(2) -r^(2)` `rArr |alpha|^(2) -1 = 8|alpha|^(2)-2` `rArr |alpha|^(2) =(1)/(7) rArr |alpha| = (1)/(sqrt(7))` |
|
| 86. |
Let complex numbers `alpha and 1/alpha` lies on circle `(x-x_0)^2(y-y_0)^2=r^2 and (x-x_0)^2+(y-y_0)^2=4r^2` respectively. If `z_0=x_0+iy_0` satisfies the equation `2|z_0|^2=r^2+2` then `|alpha|` is equal to (a) `1/sqrt2` (b) `1/2` (c) `1/sqrt7` (d) `1/3`A. `1/(sqrt(2))`B. `1/2`C. `1/sqrt(7)`D. `1/3` |
|
Answer» Correct Answer - C Formula used `|z|^2=z. bar z` and `|z_1=z_2|^2=(z_1-z_2)(barz_1-z_2)` `=|z_1|^2-z_1barz_2-z_2bar z_1+|z|^2` Here `(x-x_0)^2+(y-y_0)^2` and `(x-x_0)^2+(y-y_0)^2=4r^2` since `alpha and 1/alpha` lies on first and second respectively `therefore " "|alpha-z_0|^2=r^2 and |(1)/(bar alpha -z_0)|^2=4r^2` `rArr (alpha-z_0)(baralpha-alpha _0)=r^2` `rArr |alpha^2|-z_0 bar alpha- bar z_0 alpha+|z_0|^2=r^3` and `|1/(alpha)-z_0|^2=4r^2` `rArr 1/alpha-z_0=4r^2` `rArr 1/(|alpha|^2)-(z)/alpha-(barz_0)/(baralpha)+|z_0|^2=4r^2` Since `|alpha|=alpha.alpha` `1/(|alpha|^2)-(z_0.baralpha)/(|alpha|^2)-barz_0/(|alpha|^2).alpha+|z_0|^2=4r^2` `rArr 1-z_0baralpha-barz_0alpha+|alpha|^2|z_0|^2=4r^2|alpha|^2` On subtracting Eqs. (i) and (ii) ,we get `(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)` `rArr(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)` `rArr (|alpha|^2-1)(1-(r^2+2)/(2))=r^2(1-4|alpha|^2) ` Given `|z_0|^2=(r^2+2)/(2)` `rArr (|alpha|^2-1).((-r^2)/2)=r^2(1-4|alpha|^2)` `rArr |alpha|^2-1=-2+8|alpha|^2` `rArr |alpha|^2-1=-2+8|alpha|^2` `rArr " "7|alpha|^2=1` `therefore |alpha|=1 //sqrt(2)` |
|
| 87. |
Find the value of expression`(cospi/2+is inpi/2)(cospi/(2^2)+is inpi/(2^2)) tooo` |
|
Answer» `(cos.(pi)/(2)+ i sin.(pi)/(2))(cos.(pi)/(2^(2))+i sin.(pi)/(2^(2))) ...."to" oo` `=cos((pi)/(2)+(pi)/(2^(2)) +....)+ isin ((pi)/(2)+(pi)/(2^(2))+.....)` `=cos [(pi)/(2)(1+(1)/(2)+(1)/(2^(2))+.....)]+ isin[(pi)/(2)(1+(1)/(2)+(1)/(2^(2))+......)]` `cos[(pi)/(2)((1)/(1-(1)/(2)))] + isin [(pi)/(2)((1)/(1-(1)/(2)))] = cos pi + isin pi =-1` |
|
| 88. |
If `z=4+isqrt(7)`, then find the value of `z^3-4z^2-9z+91.` |
|
Answer» `z = 4 + I sqrt(7)` or ` x z- 4 = isqrt(7)` or ` z^(2) - 8z + 16 = - 7` or ` z^(2) - 8z + 23 = 0` Now dividing `z^(3) - 4z^(2) - 9z + 91 ` by `z^(2) - 8z + 23` we get `z^(3) - 4z^(2) - 9 z + 91 = (z^(2) - 8z + 23) (z +4) - = -1`. |
|
| 89. |
Express each of the following in the standerd from ` a + ib` (i) `(5+4i)/(4+ 5i)` (ii) ` ((1 +i)^(2))/(3-i)` (iii)` (1)/(1-cos theta + 2i sintheta)` |
|
Answer» (i) `(5+4i)/(4+5i) =(5+4i)/(4+5i)xx(4-5i)/(4-5i)` ` =((20 +20) +i(16-25))/(16-25i) =(40-9i)/(41) =(40)/(41) -(9)/(41)i` (ii) `((1+i)^(2))/(3-i) =(1+2i+i^(2))/(3-i)` `=(2i)/(3-i) =(2i)/(3-i) xx(3+i)/(3+i) = (6i+2i^(2))/(9-i)` `=(-2+6i)/(10) = -(1)/(5) + (3)/(5) i` (iii) `(1)/(1- cos thets + 2i sin theta) ` `= (1)/(1-cos theta + 2i sin theta) xx(1- cos theta - 2i sin theta)/(1- cos theta - 2i sin theta)` `=(1- cos theta - 2i sin theta)/((1- cos theta)^(2) + 4 sin theta)` `=(1-cos theta- 2i sin theta)/(1-2 cos theta + cos^(2) theta + 4 sin^(2) theta)` `= (1- cos theta - 2 i sin theta)/( 2-2 cos theta + 3 sin^(2) theta)` `= ((1-cos theta)/( 2-2 sin theta + 3 sin^(2) theta))+i((-2 sin theta)/(2-2 cos theta + 3 sin^(2)theta))` |
|
| 90. |
In an Agrad plane `z_(1),z_(2) and z_(3)` are, respectively, the vertices of an isosceles trinagle ABC with AC= BC and `/_CAB = theta`. If `z_(4)` is incentre of triangle, then The value of `(z_(2)-z_(1))^(2)tan theta tan theta //2` isA. `(z_(1) + z_(2)-2z_(3))`B. `(z_(1)+z_(2)-z_(3))(z_(1) +z_(2)-z_(4))`C. `-(z_(1) +z_(2)-2z_(3))(z_(1)+z_(2)-2z_(4))`D. `z_(4)=sqrt(z_(2)z_(3))` |
|
Answer» Correct Answer - C Keeping in mind that `tan theta = CD//AD and tan theta//2 = ID//BD`, we have `(z_(3) - (z_(1)+z_(2))/(2))/(z_(1) - (z_(1)+z_(2))/(2))=(|z_(3)-(z_(1)+z_(2))/(2)|)/(|z_(1) -(z_(1) +z_(2))/(2)|)e^(-i(pi)/(2)` `rArr (2z_(3) -z_(1) -z_(2))/(z_(1)-z_(2)) =(CD)/(AD)e^(-i(pi)/(2))" "(1)` `and (z_(4) -(z_(1)+z_(2))/(4))/(z_(2) -(z_(1)+z_(2))/(2))=(|z_(4) -(z_(1) + z_(2))/(2)|)/(|z_(2) -(z_(1) +z_(2))/(2)|)` `rArr (2z_(4) -z_(1) -z_(2))/(z_(2)-z_(1)) = (ID)/(BD) e^(-i(pi)/(2))" "(2)` Multiplying (1) and(2) we have, `(2z_(3)-z_(1) -z_(2))/(z_(1)z_(2))( 2z_(4) -z_(1) -z_(2))/(z_(2)-z_(1)) = (CD)/(AD)(ID)/(BD) = tan theta tan .(theta)/(2)` `rArr (z_(2) -z_(1))^(2) tan theta tan.(theta)/(2) = - (z_(1) +z_(2)+2z_(3))(z_(1) + z_(2)+2z_(4))` |
|
| 91. |
In an Agrad plane `z_(1),z_(2) and z_(3)` are, respectively, the vertices of an isosceles trinagle ABC with AC= BC and `/_CAB = theta`. If `z_(4)` is incentre of triangle, then The value of `(z_(4) -z_(1))^(2) (cos theta + 1) sec theta ` isA. `((z_(2)-z_(1))(z_(3)-z_(1)))/((z_(4) -z_(1)))`B. `(z_(2)-z_(1))(z_(3) -z_(1))`C. `(z_(2)-z_(1))(z_(3)-z_(1))^(2)`D. `((z_(2) -z_(1))(z_(1)-z_(3)))/((z_(4)-z_(1))^(2))` |
|
Answer» Correct Answer - B From (1), `((z_(2)-z_(1))(z_(3) -z_(1)))/(z_(2) -z_(1)) =2((AD)/(IA))^(2)((AC)/(AD)) " " (because AB = 2AD)` `rArr (z_(2) -z_(1)) (z_(3) -z_(1)) =(z_(4) -z_(1))^(2) 2 cos^(2). (theta)/(2) sec theta` `= (z_(4) -z_(1))^(2) (cos theta + 1)` |
|
| 92. |
If `a,b,c,d in R` and all the three roots of `az^3 + bz^2 + cZ + d=0` have negative real parts, thenA. `ab gt 0 `B. `bv gt 0`C. `ad gt 0`D. `bc-ad gt 0` |
|
Answer» Correct Answer - A::B::C::D Let the roots of `az^(3) + bz^(2) + cz + d = 0 ` be `z_(1) = x_(1),z_(2) = x_(2) + iy_(2)` and `z_(3) = x_(2) -iy_(2)` `therefore z_(1) + z_(2) + z_(3) = -(b)/(a)` `rArr x_(1) +2x_(2) = - (b)/(a) lt 0` `rArr ab gt 0` `z_(1)z_(2) + z_(2)z_(3) + z_(1) +z_(3) = (c)/(a)` `x_(1)(x_(2) + iy_(2)) + (x_(2)^(2) + y_(2)^(2)) + x_(1)(x_(2) -iy_(2))` ` =2x_(1)x_(2) + x_(2)^(2) + y_(2)^(2) gt 0` `rArr (c)/(0) gt0` `rArr ac gt 0` `rArr a^(2) bc gt 0 ` `bc gt 0` Also, `z_(1)z_(2)z_(3)= x_(1)(x_(2)^(2) + y_(2)^(2)) = -(d)/(a)` `rArr ad gt 0` Futher,`-(bc)/(a^(2)) = (-(b)/(a))((c)/(a))` `= (x_(1)+ 2x_(2)) (2x_(1)x_(2) + x_(2)^(2) + y_(2)^(2))` `=x_(1)(x_(2)^(2) + y^(2)_(2)) + 2x_(1)^(2) x_(2) + 2x_(1)x_(2)^(2) + 2x_(2) (x_(2)^(2) + y_(2)^(2))` `lt x_(1)(x_(2)^(2) + y_(2)^(2))` `rArr -(bc)/(a^(2)) lt - (d)/(a)` `rArr bc gt ad` |
|
| 93. |
Number of solutions of the equation `z^3+[3(barz)^2]/|z|=0` where z is a complex number isA. 2B. 3C. 6D. 5 |
|
Answer» Correct Answer - D Given, `z^(3)+(3(barz)^(2))/(|z|)=0` Let `z=re^(itheta)` `rArr r^(3)e^(i3theta)+3re^(-2theta)=0` Since r cannot be zero, so `re^(i5theta)=-3` Which will hold for r = 3 and five values of `theta`. Thus, there are five solution. |
|
| 94. |
Find the real numbers `x and y ,`if `(x-i y)(3+5i)`is the conjugate of `-6-24 i` |
|
Answer» Correct Answer - ` x = 3, y=-3` Let `z = (x-iy) (3+5i)` `=3x + 5xi - 3yi - 5yi^(2)` `= (3x + 5y)+i(5x -3y)` `therefore barz = (3x + 5y) - i(5x - 3y)` It is given that ,`barz = - 6- 24i` Equating real and imaginary parts, we get ` 3x + 5y = -6` `5x -3y = 24` Sloving we get `x = 3, y = -3` |
|
| 95. |
If `Z^5` is a non-real complex number, then find the minimum value of `(Imz^5)/(Im^5z)` |
|
Answer» Correct Answer - `-4` Let z= a+ ib, `b ne 0` where Im Z = b `therefore " " Z^(5) = (a+ib)^(5)` `= a^(5) + 5a^(4) bi + 10^(3)b^(2)i^(2) + 10a^(2)b^(3)i^(2) + 5ab^(4)i^(4) + i^(5)b^(5)` `therefore" " Lm z^(5)=5a^(4)b- 10a^(2)b^(3) + b^(5)` `y= (ImZ^(5))/((ImZ^(5)))=((a)/(5))^(2)-10((a)/(b))^(2)+1` `= 5[((a)/(b))^(4)-2((a)/(b))^(2)+(1)/(5)]` `=5[(((a)/(b))^(2)-1)^(2) +(1)/(5)-1]` `=5(((a)/(b))^(2)-1)^(2)=4` So, `y_("min")= -4` |
|
| 96. |
The complex number, `z=((-sqrt(3)+3i)(1-i))/((3+sqrt(3)i)(i)(sqrt(3)+sqrt(3)i))`A. lies on real axisB. lies on imaginary axisC. lies in first quadrantD. lies in second quadrant |
|
Answer» Correct Answer - B `(b)` `((-sqrt(3)+3i)(1-i))/((3+sqrt(3)i)(i)(sqrt(3)+sqrt(3)i))` `=(sqrt(3)(-1+sqrt(3)i)(1-i))/(3(sqrt(3)+i)(i)(1+i))` `=((-1+sqrt(3)i)(1-i))/(sqrt(3)(-1+sqrt(3)i)(1+i))` `=((1-i)^(2))/(2sqrt(3))=(-2i)/(2sqrt(3))`, which lies on imaginergy axis |
|
| 97. |
The number of complex numbers `z`satisfying `|z-3-i|=|z-9-i|a n d|z-3+3i|=3`area. oneb. twoc. fourd. none of theseA. oneB. twoC. fourD. none of these |
|
Answer» Correct Answer - A Let `z = x +iy`. Then, `|z-3-i|=|z-9-i|` `rArr sqrt((x-3)^(2)+(y -1)^(2))=sqrt((x-9)^(2)+(y-1)^(2))` `rArr x = 6` `|z-3+3i|=3` `rArr sqrt((x-3)^(2)+(y+3)^(2))=3` For `x =6, y=-3`. `therefore z =6-3i` |
|
| 98. |
Simplify:`(sqrt(5+12 i)+sqrt(5-12 i))/(sqrt(5+12 i)-sqrt(5-12 i))` |
|
Answer» Correct Answer - `-(3)/(2)i` `((sqrt(5+12i)+sqrt(5-12i))(sqrt(5+12i)+sqrt(5-12i)))/((sqrt(5+12i)-sqrt(5-12i))(sqrt(5+2i)+sqrt(5-12i)))` `(5+12i+5-12i+2sqrt(5+12i)sqrt(5-12i))/(5+12i-5+12i)` `=(10+2xx13)/(24i)` ` = -(3)/(2)i` |
|
| 99. |
If `z=2-3i`show that `z^2-4z+13=0`and hence find the value of `4z^3-3z^2+169.` |
|
Answer» `z = 2-3i` Now, `z^2-4z+13 = (2-3i)^2-4(2-3i)+13` `=4+9i^2-12i-8+12i+13` `=4-9+5 = 0...[As i^2 = -1]` `:. z^2-4z+13 = 0` Now, `4z^3-3z^2+169 = 4z^3-16z^2+13z^2+52z-52z+169` `=4z^3-16z^2+52z+13z^2-52z+169` `=4z(z^2-4z+13) + 13(z^2-4z+13) ` `=4z(0)+13(0) = 0` `:. 4z^3-3z^2+169 = 0.` |
|
| 100. |
If `x_n`=cos`pi/(2^n)`+isin`pi/(2^n)`, prove that `x_1x_2x_3 x_oo=-1.` |
|
Answer» Here, `x_n = cos(pi/(2^n))+isin(pi/(2^n))` `:.L.H.S. = x_1x_2x_3...x_oo` `= (cos(pi/(2))+isin(pi/(2)))(cos(pi/(2^2))+isin(pi/(2^2)))(cos(pi/(2^3))+isin(pi/(2^3)))...oo` `=cos(pi/2+pi/2^2+pi/2^3+...oo)+isin(pi/2+pi/2^2+pi/2^3+...oo)` `= cos((pi/2)/(1-1/2))+isin((pi/2)/(1-1/2))` `=cospi+isinpi` `= -1+0 = -1 = R.H.S.` |
|