Mathematics Advanced Discussion On Complex Number For IIT JEE Part I :

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Advanced Discussion On Complex Number For IIT JEE Part I :

Basic facts :

(i) `N subset W subset I subset Q subset R subset Z`

(ii) `sqrt(-3) sqrt(-4) ne sqrt((-3)(-4))` from `sqrt a sqrt b =sqrt(ab)` at least one of a and b should be non-negative.

(iii) `x^2+y^2=0 => x =0, y=0 ` `forall x,y in R`

but `z_1^2+z_2^2 =0 ne z_1 =0 , z_2=0` `forall z_1, z_2 in C`

`i.e. (1+i)^2 +(1-i)^2=0 = 1-1+2i + 1-1-2i = 0`

Q 1426723671

Consider the following statements :

`S_1 :-8=2i xx 4i = sqrt (-4) xx sqrt(-16)`

`S_2 : sqrt(64)=8`

`S_3 : sqrt((-4) xx (-16)) =sqrt (64)`

`S_4 : = sqrt (-4) xx sqrt(-16)= sqrt((-4) xx (-16))`

of these statements, the incorrect one is

(A)

`S_1` only

(B)

`S_2` only

(C)

`S_3` only

(D)

`S_4` only

Solution:

If `a < 0` and `b < 0`

Then `sqrt a sqrt b=-sqrt(ab)`

`:. sqrt(-4) xx sqrt (-16)=-sqrt (64)=-8`

Hence, `S_4` is incorrect.

`

Correct Answer is `=>` (D) `S_4` only

Exponential Form and Euler's Form of Complex Number

`z = r(cos theta + i sin theta) = re^(i theta)`

`z = re^(i theta) ` is called exponential form of the complex number.

where `r` is modulus of `z` and `theta` is amplitude of `z`.

Now If `r = 1` in exponential form

then `z = e^(itheta) = costheta + isintheta ` (is known as Euler's identity.)

`z = e^(itheta)` is known as euler's form of complex number

Now, `barz = e^(-itheta) = costheta - isintheta `

Here, `|z| = |barz| = 1`

Here, `cos theta + i sin theta =e ^(i theta) forall theta`..................(1)

Replacing `i` by `-i` we get,

`cos theta-i sin theta=e^(- i theta) forall theta`...................(2)

Adding ( 1) and (2)

`cos theta=(e^(i theta)+ e^(-i theta))/2` which is purely imaginary.

and if `theta, phi in R` and `i= sqrt(-1) ` then

(i) `e^(i theta)+ e^(i phi) = e^(i (theta+ phi)/2)* 2 cos ((theta- phi)/2)`

`:. | e^(i theta)+ e^(i phi) |= 2| cos ((theta- phi)/2)|`

and `arg (e^(i theta)+ e^(i phi))) =((theta+ phi)/2)`

(ii) `e^(i theta)- e^(i phi) = e^(i(theta+phi)/2) * 2 i sin ((theta- phi)/2)`

`:. | e^(i theta) - e^(i phi)|= 2| sin ((theta- phi)/2)|`

and `arg(e^(i theta)- e^(i phi))=(theta+ phi)/2 + pi/2`

Q 2610456319

Number of solution of the equation `z^3 + (3 (bar z)^2)/(|z|) = 0` where `z ` is a complex number is

Solution:

`z^3 + (3 (bar z)^2)/(|z|) = 0` Let `z = re^(i theta)`

`=> r^3 e^(i 3 theta) + 3 re^( -2 theta) = 0`

Since '`r`' cannot be zero

` => r^2 e^(i 5 theta) = -3 ` which will hold for

`r = sqrt3` and `5` distinct values of '`theta`'

Thus there are five solution.

Q 2600467318

Let ` f_p(alpha) = e^((i alpha)/p^2) , e^((2 i alpha)/p^2) ....... e^((i alpha)/p) p in N ` (where `i = sqrt(-1)` , then find the value of ` | lim_(n -> oo) f_n (pi) |`

Solution:

`f_p(alpha) = e^(((i alpha)/p^2) (1 + 2 + ... p) ) = e^ (((i alpha)/2) (1 + 1/p))`

` lim_(n -> oo) f_n (pi) = lim_(n -> oo) e^ (((i alpha)/2) (1 + 1/n)) = e^((i alpha)/2)`

` | lim_(n -> oo) f_n (pi) | = | e^((i alpha)/2) | = 1`Ans.

Q 2651401324

If `|z| =1` and `Arg (z + z omega) = pi/2` , then one of the arguments of `z` is `pi/6` (where `omega` is a non-real cube root of unity) what is is the other value of argument of `z`

(A)

True

(B)

False

Solution:

`z+ z omega =e^(i theta)( 1+ omega)= e^(i theta)(1+ (-1 pm sqrt 3i)/2)=e^(i theta) ((1 pm sqrt 3 i)/2)`

`=e^(i theta) e ^(pm i pi/3)= e^((theta pm pi/3))`

`:. theta pm pi/3 =pi/2 `

`:. theta=pi/2 pm pi/3 =pi/6 , (5 pi)/6`

Correct Answer is `=>` (A)

Q 2651301224

If `z` is a complex number such that `| z | = 1` and `arg (z/(bar z))=pi/2` then , `Re(z)= pm 1/sqrt 2`

(A)

True

(B)

False

Solution:

Since `| z | = 1`, therefore, let `z = e^(i theta)`

`Arg (z/(bar z))= pi/2 => cos 2 theta=0 => theta =pi/4 , (5 pi)/4`

`:. z= 1/sqrt 2 + 1/sqrt 2 i, (-1)/(sqrt 2) -1/(sqrt 2) i` i.e., `Re(z)= pm 1/sqrt 2`

Correct Answer is `=>` (A)

Q 2681201127

Number of solutions of the equation `z^3 = bar z i|z|` are `5`

(A)

True

(B)

False

Solution:

`z^3= bar z i |z| => | z^3| = | bar z| | i | |z| `

`=> |z| =1` or `|z|=0`

Thus `z = 0` is a solution

if `|z|=1, ` Let `z=e^(i theta)` then `e^ (3 i theta)=ie^(-i theta) `

`=> e^(4 i theta) =i`

`=> 4 theta =pi/2 ,( 5pi)/2 , (9 pi)/2 , (13 pi)/2`

`:. theta =pi/8 , (5 pi)/8 , (9 pi)/8 , (13 pi)/8` are solutions.

`:.` In all there are 5 solutions

Correct Answer is `=>` (A)

Properties of Argument of complex number :

(i) `arg(z_1z_2) = arg(z_1) + arg(z_2) + 2kpi, k in I`

In general , `arg (z_1,z_2,z_3........z_n)`

`=arg(z_1) + arg(z_2) + arg(z_3)+ ..............+ agr(z_n) + 2k pi , k in I`

`(ii) arg(z_1/z_2) = arg(z_1) - arg(z_2) + 2k pi, k in I`

where, proper value of k must be chosen, so that RHS lies in `(-pi,pi)` .

`(v) ` If `arg(z_2/z_1) = theta.` then `arg(z_1/z_2) = 2npi - theta , ` where `n in I`

`(vi) arg(bar z) = - arg(z)`

Q 2600556418

The point of intersection the curves arg `(z - i + 2) = pi/6` & `arg (z + 4 - 3i) = - pi/4` is given by

(A)

`(- 2 + i)`

(B)

`2- i`

(C)

`2 + i`

(D)

None of these

Solution:

`arg (z - i + 2) = pi/6 => tan pi/6 = (y - 1)/(x +2)`

`=> x - sqrt3 y =- (sqrt3 + 2) , x > -2, y > 1`

`arg (z + 4 - 3i) = - pi/4 => tan (- pi/4) = (y-3)/(x+4)`

`=> y + x = - 1 , x > - 4, y < 3`

so, there is no point of intersection.

Correct Answer is `=>` (D) None of these

Q 2630756612

The principal argument of the complex number ` ( (1 + i)^5 (1 + sqrt3 i)^2)/(- 2i( - sqrt3 + i))` is

(A)

`(19 pi)/(12)`

(B)

` - (7 pi)/(12)`

(C)

`-(5 pi)/(12)`

(D)

`(5 pi)/(12)`

Solution:

` ( (1 + i)^5 (1 + sqrt3 i)^2)/(- 2i( - sqrt3 + i)) = ( (sqrt2)^5 ( 1/sqrt2 + i/sqrt2)^5 .2^2 ( 1/2 + sqrt3/2 i)^2)/( 2 i 2 ( sqrt3/2 - i/2))`

`:.` argument ` = (5pi)/4 + (2pi)/3 - pi/2 + pi/6 = (19 pi)/(12)`

`:. ` principal argument is ` - (5 pi)/(12)`

Correct Answer is `=>` (C) `-(5 pi)/(12)`

Q 2680856717

If a complex number `z` satisfies `| 2z + 10 + 10i | <= 5 sqrt3 - 5` then the least principal argument of z is

(A)

` (11 pi)/(12)`

(B)

`- (2 pi)/3`

(C)

` -(5 pi)/6`

(D)

` (3 pi)/4`

Solution:

Point `B` has least principal argument

`AB = ( 5(sqrt3 -1))/2`

`OA = 5sqrt2`

`angle AOB = pi/(12)`

`:. Arg (z) = - (5 pi)/6`

Correct Answer is `=>` (C) ` -(5 pi)/6`

Rotation of Complex Number :

f `z` and `z'` are two complex numbers then argument of `z/(z')` is the angle through which `Oz'` must be turned in order that it may lie along `Oz` .

`z/(z')=(|z|e^(i theta))/(|z'| e^(i theta'))=(|z|)/(|z'|)e^(i(theta-theta'))=(|z|)/(|z'|)e^(i alpha)`

In general, let `z_1,z_2, z_3` be the three vertices of a triangle `ABC` described in the counter-clock wise sense. Draw `OP` and `OQ` parallel and equal to `AB` and `AC` respectively. Then the point `P` is `z_2 - z_1` and `Q` is `z_3 - z_1` and

`(z_3-z_1)/(z_2-z_1)=(OQ)/(OP) (cos alpha+i sin alpha)`

`(CA)/(BA) . e^(i alpha)=(|z_3-z_1|)/(|z_2-z_1|) . e^(i alpha)`

Note that arg. `(z_3 - z_1) - arg(z_2 - z_1) = a` is the angle through which ` OP` must be rotated in the anti-clockwise direction so that it conciders with `OQ` .

Here we can write `(|z_3-z_1|)/(z_2-z_1)=(|z_3-z_1|)/(|z_2-z_1|) . e^(-i(2 pi-alpha))` also. ln case we are rotating `OP` in clockwise direction by an angle `(2 pi - alpha)`. Since the rotation is in clockwise direction, we are taking negative sign with angle
`(2 pi - alpha)`.

Q 2680280117

Consider a square `ABCD` such that `z _1, z _2, z_ 3` and `z_4` represent its vertices `A, B, C` and `D` respectively. Express `'z_3'` and `'z_4'` in terms of `z_1` and `z_2`.

Solution:

Consider the rotation of `A B` about `A` through an angle `pi/4` We get

`(z_3-z_1)/(z_2-z_1) =|(z_3-z_1)/(z_2-z_1)| e^(i pi//4)`

`= sqrt 2 (cos (pi/4) + i sin (pi/4))=> z_3 =z_1 + (z_2 -z_1) (1+i)`

Similarly, `(z_4-z_1)/(z_2-z_1) = |(z_4-z_1)/(z_2-z_1)| | (z_4-z_1)/(z_2-z_1)| =i`

`=> z_4 =z_1 + i(z_2 -z_1)`

De-Moivre's Theorem :

`text(Statements)`

`(i)` If `theta_1 , theta_2 , theta_3 , ........ , theta_n in R` and `i - sqrt(-1),` then
`(cos theta_1+ i sin theta_1)(cos theta_2+ i sin theta_2)(cos theta_3+ i sin theta_3)............(cos theta_n+ i sin theta_n)`
`= cos (theta_1+ theta_2+theta_3+.....+ theta_n) + isin(theta_1+theta_2+theta_3+.......+theta_n)`

`(ii)` If `theta in R , n in I` (set of integers) and `i=sqrt(-1),` then
`(cos theta + i sin theta)^n = cos n theta+ isin n theta`

`text(Important Result)`

`1. (cos theta - i sin theta)^n = cos n theta - i sin n theta , AA n in I`

`2. (sin theta - i cos theta)^n =(i)^n (cos n theta - i sin ntheta) , AA n in I`

`3. (sin theta - i cos theta)^n =(-i)^n (cos n theta + i sin ntheta) , AA n in I`

`4. (cos theta + i sin phi)^n ne cos n theta + i sin phi , AA n in I`
[here, `theta ne phi` De-Moivre's theorem is not applicable]

Note : The main application of DMT is to find roots of complex number.

Q 2406445378

If `x ^2- 2x cos theta + 1 = 0`, then `x^(2n) -2x^n cos n theta+1` is equal to
UPSEE 2010

(A)

`cos 2 n theta`

(B)

`sin 2 n theta`

(C)

`0`

(D)

`R-{0}`

Solution:

From given relation, we have

`x = cos theta ± i sin theta`

take `x = cos theta + i sin theta`

`x^n = cos ntheta + i sin n theta`

and `1/ x^n =cos n theta- i sin n theta`

`:. x^n + (1// x^n) = 2cos n theta`

`:. 2x^(2n) cos ntheta + 1 = 0`

Correct Answer is `=>` (C) `0`

Q 2404580458

If `x_a = cos (pi/4^n) + i sin (pi/4^n)`, then

`x_1 * x_2 * x_3 ................oo` is
UPSEE 2012

(A)

`(1+ i sqrt (3) )/2`

(B)

`(-1 + i sqrt (3) )/2`

(C)

`(1- i sqrt (3) )/2`

(D)

`( -1 -i sqrt (3))/2`

Solution:

`x_1 * x_2 * x_3 ........oo`

`= [cos (pi/4)+ i sin (pi/4) ] [cos (pi/4^2) + i sin (pi/4^2) ]`

` * [cos (pi/4^3) + i sin (pi/4^3) ] ....oo`

`= cos (pi/4 + pi/4^2 + pi/4^3 + .........oo)`

`+i sin (pi/4 + pi/4^2 + pi/4^3 + .............oo)`

`=cos ((pi/4)/(1-1/4)) + i sin ((pi/4)/(1-1/4))`

`=cos (pi/3) + i sin (pi/3) = (1+ sqrt (3) i)/2`

Correct Answer is `=>` (A) `(1+ i sqrt (3) )/2`

Cube Roots of unity :

`z^3-1=0=>z=(1)^(1//3)=(cos theta +i sin theta)^(1//3)=(cos 2 m pi+i sin 2 m pi)^(1//3)`

`= cos frac (2 m pi)(3)+i sin frac(2 m pi)(3),m-=0,1,2`

`m=0, z_1= cos 0+i sin 0=1`

`m=1, z_2=cos frac (2 pi )(3)+i sin frac (2 pi)(3)= -1/2+i (sqrt 3)/2= omega`

`m=2, z_3=cos frac(4 pi)(3), +i sin frac (4 pi)(3)=(-1)/2-i (sqrt 3)/2=omega^2`

`text(Important Results :)`

`1. bar omega = omega^2 , \ \ (bar omega)^2 = omega`

`2. sqrt(omega) = pm omega^2, \ \ sqrt(omega^2) = pm omega`

`3. |omega| = |omega^2| = 1`

Q 2610191919

If `omega` is an imaginary fifth root of unity, then `log_2 |1+ omega +omega^2+omega^3-1/omega|=1`

(A)

True

(B)

False

Solution:

`log_2 |1+ omega+omega^2+omega^3 -omega^4|= log_2|- 2 omega^4|= log_2 2 =1`

`(|omega^4| =1 ` )

Correct Answer is `=>` (A)

Q 2680156917

Sum of common roots of the equations `z^3 + 2z^2 + 2z + 1 = 0` and `z^(97) + z^(29) + 1 = 0` is equal to

(A)

`0`

(B)

`-1`

(C)

`1`

(D)

None

Solution:

`z^3 + 2z^2 + 2z + 1 = (z^3 + 1) + 2z (z + 1)`

`= (z + 1) (z^2 + z + 1) = 0`

`=> z = -1, omega ,omega ^ 2`

`- 1` does not satisfy `z^(97) + z^(29) + 1 = 0`

but `omega ` and `omega ^2` satisfy and `omega + omega ^2 = - 1`

Correct Answer is `=>` (B) `-1`

Q 2640267113

If `z_1 = a + i b` and `z_2 = c + id` (where `i = sqrt(-1)`) are two complex numbers such that `|z_1| = |z_2| = 1` and `Re (z_1 z_2) = 0`, then the pair of complex number `omega_1 = a + ic` and `omega_2 = b + id` satisfy

(A)

`|omega_1| = 1`

(B)

`|omega_2| = 1`

(C)

`|omega_1 bar omega_2 | =1`

(D)

`Re |bar omega_1 omega_2 | =0`

Correct Answer is `=>` (A)

Q 2680567417

Assertion : If `x+1/x =1` and `p=x^(4000) + 1/(x^(4000))` and `q` be the digit at unit place in the number `2^(2^n)+1, n in N` and `n > 1` then, the value of `p+q=8`

Reason : `omega, omega^2` are the roots of `x+ 1/x=-1` , then `x^2 + 1/x^2=-1 , x^3 + 1/(x^3)=2`

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Solution:

`x+ 1/x=1`

`=> x^2-x+1=0`

`:. x=- omega, -omega^2`

Now for `x=- omega, P= w^(4000)+ 1/(omega^(4000))= omega+ 1/omega=-1`

Similarly for `x=-omega^2` also `p=-1`

for `n > 1, 2^n =4k`

`:. 2^(2^n) =2^(4k)=(16)^k=a` number with last digit `= 6`

`=> q=6+1=7`

Hence, `p+ q =-1 + 7=6`

Correct Answer is `=>` (D)

Q 2640267113

(A)

`|omega_1| = 1`

(B)

`|omega_2| = 1`

(C)

`|omega_1 bar omega_2 | =1`

(D)

`Re |bar omega_1 omega_2 | =0`

Correct Answer is `=>` (A)

`n , n^(th)` root of unity :

`z^n-1=0 =>\ \ \ \ \ \ z=(1)^(1/n)=(cos0+isin0)^(1/n)=(cos2mpi+i sin 2mpi)^(1/n)`

`=>cos((2mpi)/n)+i sin ((2mpi)/n) ,\ \ \ \ m=0, 1, 2 , 3 ,............,(n-1)`

`m = 0, z_1 = 1`

`m=1 , z_2= cos((2pi)/n)+i sin ((2pi)/n)= e^(i(2pi)/n)=alpha(Let)`

`m=2 , z_3= cos((4pi)/n)+i sin ((4pi)/n)= e^(i(4pi)/n)=alpha^2(Let)`
` .`
` .`
`. `
` .`
` .`
` .`
`. `
` .`
`m= (n - 1), z_n = cos((2(n-1)pi)/n)+i sin ((2(n-1)pi)/n)= e^(i((2(n-1)pi)/n))=alpha^(n-1)(Let)`

`text(Important Results : )`

`1.` If `1, alpha_1, alpha_2,alpha_3,..........,alpha_(n-1)` nth roots of unity, then

`(1)^p + (alpha_1)^p + (alpha_2)^p + ......+ (alpha_(n-1))^p = { tt[(0, text{if pis not an integral multiple of n}),(n,text{1f p is an mtegral multiple of n})]`

`2. (1 + alpha_1)(1-alpha_2).............(1- alpha_(n-1)) = { tt[(0, text{if n is even}),(1,text{if n is odd})]`

`3.(1-alpha_1)(1-alpha_2)..........(1-alpha_(n-1)) = n`

`4. z^n -1 = (z-1)(z+1) Pi_(r=1)^(n-2)//2(z^2 - 2z cos (2rpi)/n + 1)` if 'n' is even

`5.z^n +1 = Pi_(r=1)^((n-2)//2)(z^2-2z cos(((2r+1)pi)/pi)+1),` if n even

`6.z^n +1 = Pi_(r=0)^((n-3)//2)(z^2-2z cos(((2r+1)pi)/pi)+1),` if 'n' odd

`text(The Sum of the Following Series Should be Remembered : )`

`(i) cos theta + cos 2 theta + cos 3 theta + .........+ cos n theta = (sin((n theta)/2))/(sin(( theta)/2)) . cos[((n+1)/2)theta]`

`(ii) sin theta + sin 2 theta + sin 3 theta + .........+ cos n theta = (sin((n theta)/2))/(sin(( theta)/2)) . sin[((n+1)/2)theta]`

Q 2660556415

If ` 1 , alpha_1 , alpha_2 ,alpha_3 .............` and `alpha_8` are nine, ninth roots of unity (taken in counter-clockwise sequence) then `|2 - alpha_1) (2 - alpha_3) (2 - alpha_5) (2 - alpha_7) | ` is equal to

(A)

` sqrt(255)`

(B)

` sqrt(511)`

(C)

` sqrt(1023)`

(D)

`15`

Correct Answer is `=>` (B) ` sqrt(511)`

Q 2661101025

The product of all the fifth roots of `-1` is equal to `-1`

(A)

True

(B)

False

Solution:

`z^5+1=0`

`:.` product of all the roots `= (-1 )^5 * 1 = -1`

Correct Answer is `=>` (A)

JEE WarmUp Examples

Solved examples

Q 2543545443

Expression of `( 1 + i)^(-i)`, (where, `i = sqrt(-1)`) in the form `A+ iB`.

(A)

`e^(pi//4) cos (log_e (1/sqrt2 )) `

(B)

`e^(pi//4) cos (log_e (1/sqrt2 )) - i e^(pi//4) sin (log_e ( 1/sqrt2))`

(C)

`e^(pi//4) cos (log_e (1/sqrt2 )) + i e^(pi//4) sin (log_e ( 1/sqrt2))`

(D)

None of these

Solution:

Let `A + iB = ( 1 + i)^(- i)`

On taking logarithm both sides, we get

`log_e (A + iB) = - i log_e (1 + i)`

`= - i log_e ( sqrt2 ( 1/sqrt2 + i/sqrt2))`

`= - i log_e ( sqrt2 (cos pi/4 + i sin pi/4))`

`= - i log_e ( sqrt2 e^(ipi//4)) = - i (log_e sqrt2 + log_e e^(ipi//4))`

`= - i ( 1/2 log_e 2 + (i pi)/4 ) = - i/2 log_e2 + pi/4`

`:. A + iB = e^(- i/2 log_e 2 + pi/4) = e^(pi//4) . e^(i log_e 2^(- 1//2) )`

` = e^(pi//4) . (cos (log_e 2^(- 1//2)) + i sin (log_e 2^(-1//2))`

`= e^(pi//4) cos (log_e (1/sqrt2 )) + i e^(pi//4) sin (log_e ( 1/sqrt2))`

Correct Answer is `=>` (C) `e^(pi//4) cos (log_e (1/sqrt2 )) + i e^(pi//4) sin (log_e ( 1/sqrt2))`

Q 2513001840

Find the value of ` sum_(r = 1)^(4n + 7) i^r ` (where, `i = sqrt(-1)` ).

(A)

(B)

-1

(C)

(D)

None of these

Solution:

` sum_(r = 1)^(4n + 7) i^r = i^1 + i^2 + i^3 + sum_(r = 4)^(4n + 7) i^r = i - 1 - i + sum_(r = 1)^(4n + 4) i^(r+3)`

`= - 1 + 0` [`(n + 1)` sets of four consecutive powers of `i`]

`= - 1`

Correct Answer is `=>` (B) -1

Q 1177323286

If the complex number z satisfies `| iz + 2 | = Im(z)`, then

(This question may have multiple correct answers)

(A) Maximum value of `| z |` is `sqrt3`

(B) Minimum value of `argz` is `π/4`

(D) Maximum value of `argz` is `(3π)/4`

Solution:

`| iz + 2 | = Im (z) `

`⇒ | z - 2i | = Im (z)`

⇒ Distance of z from `2i =` Distance of z from real axis

⇒ Locus of z is parabola having focus at `(0,2)` and directrix at x - axis.

So its Cartesian Equation is

`x^2 = 4 ( y - 1 )`

Clearly minimum value of `| z | = OV = 1` ( V is the vertex of parabola )

Maximum value of | z | is ∞.

Also point A and B respectively represent minimum and maximum values of argument .

[OA and OB represent tangents from origin Whose equation are `y=±x`]

Correct Answer is `=>` (B)

Q 2543045843

Find the square roots of the following :
(i) 4 + 3i
(ii) - 5 + 12i
(iii) -8- 15i
(iv) 7- 24i
(where, `i = sqrt(-1)`)

Solution:

(i) Let `z = 4 + 3i`

`:. | z | = 5 , Re (z) = 4, Im (z) = 3 > 0`

` ∵ sqrtz = pm ( sqrt((| z | + Re (z))/2) + i sqrt((| z | - Re (z))/2) )`

`:. sqrt((4 + 3 i)) = pm ( sqrt((5 + 4)/2) + i sqrt((5 - 4)/2) ) = pm ( ( 3 + i)/sqrt2)`

(ii) Let `z = -5 + 12i`

`:. |z| = 13, Re (z) = -5, Im (z) = 12 > 0`

`∵ sqrtz = ± ( sqrt((| z | + Re (z))/2) + i sqrt((| z | - Re (z))/2) )`

`:. sqrt(-5 + 12i) = ± ( sqrt((13 - 5)/2) + i sqrt((13 + 5)/2) ) = pm (2 + 3i)`

(iii) Let `z = - 8 - 15i`

`:. | z | = 17, Re (z) =- 8, Im (z) = -15 < 0`

` ∵ sqrtz = ± ( sqrt((| z | + Re (z))/2) - i sqrt((| z | - Re (z))/2) )`

`:. sqrt(-8 - 15i) = pm ( sqrt ((17 - 8)/2) - i sqrt ((17 + 8)/2) ) = pm ((3 - 5i)/sqrt2)`

(iv) Let `z = 7 - 24i`

`:. | z | = 25, Re (z) = 7, Im (z) = - 24 < 0`

` ∵ sqrtz = ± ( sqrt((| z | + Re (z))/2) - i sqrt((| z | - Re (z))/2) )`

`:. sqrt(7- 24i) = ± ( sqrt ((25 + 7)/2) - i sqrt ((25 - 7)/2) ) = pm ( 4 - 3i)`

Q 2563078845

The area of the triangle on the argand plane formed by the complex numbers `z, iz` and `z + iz` is `1/2 | z |^2` where `i = sqrt(-1)`.

(A)

`1/2 | z |^2`

(B)

`1/2 | z |`

(C)

`1/2`

(D)

None of these

Solution:

Required area `= 1/4 | (z , bar z , 1),(iz ,bar (iz) , 1),(z + iz , bar(z + iz) , 1) | = 1/4 | (z , bar z , 1),(iz ,bar (iz) , 1),(z + iz , bar(z + iz) , 1) |`

` = 1/4 | (z , bar z , 1),(iz , -i bar z , 1),(z + iz , barz - i bar z , 1) |`

On applying `R_3 -> R_3 - (R_1 + R_2)`, we get

Area ` = 1/4 | (z , bar z , 1),(iz , -i bar z , 1),(0 , 0 , -1)| = 1/4 | (-1) (- iz bar z - iz bar z) |`

` = 1/4 | 2 i z bar z| = 1/2 | i | | z bar z| = 1/2 | z |^2`

Correct Answer is `=>` (A) `1/2 | z |^2`

Q 2362801735

If `z + 1 = i sqrt(3)` and `n` is a positive integer

but not a multiple of `3`, then `z^(2n) + 2^n z^n` is equal to
BITSAT Mock

(A)

`0`

(B)

`-1`

(C)

`2^(2n)`

(D)

`-2^(2n)`

Solution:

`z = - 1 + i sqrt(3) = 2 omega`, where `omega` is a cube root of unity.

`=> z^(2n) + 2^n z^n = 2^(2n) (omega^(2n) + omega^n)`

`= 2^(2n) (omega^2 + omega)`

irrespective of whether `n` is of the

form `3m + 1 ` or `3m + 2`

`= - 2^(2n)`

Correct Answer is `=>` (D) `-2^(2n)`

Q 1415791669

If `w` is a complex cube root of unity, then `cos [ { (1- w) (1- w^2) + ... + (10- w) (10- w^ 2
)} pi/900]`

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`sqrt 3/2`

Solution:

`cos [{1-w)(1-w^2)+.....(10-w)(10-w^2)} pi/900]`

`=cos [{sum_(r=1)^(10)(r-w)(r-w^2)} pi/900]`

`=cos [ { sum_(r=1)^(10)(r^2+r+1)} pi/900]`

`=cos[((10 xx 11 xx 21)/6 +(10 xx 11)/2) pi/900]=cos (pi/2)=0`

Correct Answer is `=>` (B) `0`

Q 2251680524

Let z be a complex number satisfying `z^2 + 2alphaz + 1 = 0` where `alpha` is a parameter which can take any real value.
The roots of this equation lie on a certain circle if
JEE Advanced Paper 2

(A)

`-1 < alpha < 1`

(B)

`alpha > 1`

(C)

`alpha < 1`

(D)

None of these

Solution:

`-16 z = - alpha pm sqrt(alpha^2 -1)`

Case I when `-1 < alpha < 1`

We get `alpha^2 < 1`

` z = - alpha pm i sqrt( 1- alpha ^2)`

or ` x = - alpha => y^2 = 1- x^2`

or `x^2 + y^2 = 1`

Case II `alpha > 1`

`alpha^2 - 1 > 0`

`z = - alpha pm sqrt(alpha^2 - 1)`

or `x = - alpha pm sqrt(alpha^2 -1) , y =0`

Roots are `(- alpha + sqrt(alpha^2 -1) , 0) , ( - alpha , - sqrt( alpha^2 -1), 0)`
One root lies inside the unit circle and other will lies outside the unit circle
Case III where `alpha` is very large then

`z = - alpha - sqrt(a^2 - 1) `

`= - 2 alpha`

`z = - alpha + sqrt(alpha^2 - 1) = -1/(2 alpha)`

Correct Answer is `=>` (A) `-1 < alpha < 1`

Q 2680567417

Assertion : If `x+1/x =1` and `p=x^(4000) + 1/(x^(4000))` and `q` be the digit at unit place in the number `2^(2^n)+1, n in N` and `n > 1` then, the value of `p+q=8`

Reason : `omega, omega^2` are the roots of `x+ 1/x=-1` , then `x^2 + 1/x^2=-1 , x^3 + 1/(x^3)=2`

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Correct Answer is `=>` (D)

Q 2543480343

If `| z | >= 3`, then determine the least value of ` | z + 1/z|` .

(A)

1/3

(B)

8/3

(C)

2/3

(D)

Solution:

`∵ | z + 1/z| >= | |z | - | 1/z | | = | | z | - 1/(|z|) |`

` ∵ |z | >= 3 => 1/(|z|) <= 1/3 ` or ` - 1/(|z|) >= - 1/3`

`:. |z | - 1/(|z|) >= 3 - 1/3 = 8/3`

` => |z | - 1/(|z|) >= 8/3`

or `| |z | - 1/(|z|) | >= 8/3`

From Eqs. (i) and (ii), we get

` | z + 1/z | >= 8/3`

`:.` Least value of ` | z + 1/z | ` is ` 8/3`.

Correct Answer is `=>` (B) 8/3

Q 2543067843

Let `z_1` and `z_2` be roots of the equation `z^2 + pz + q = 0`, where the coefficients `p` and `q` may be complex numbers. Let `A` and `B` represent `z_1` and `z_2` in the complex plane. If `angle AOB = alpha != 0` and `OA = OB`, where `O` is the origin, then `p^2 = ?.`

(A)

`0`

(B)

`4`

(C)

`-4 q cos^2 (alpha// 2)`

(D)

`4 q cos^2 (alpha// 2)`

Solution:

Clearly, `vec(OB)` is obtained by rotating `vec(OA)` through angle `alpha`

`:. vec(OB) = vec (OA) e^(i alpha)`

`=> z_2 = z_1 e^(i alpha)`

`=> z_2/z_1 = e^(i alpha)` ......(i)

or ` z_2/z_1 + 1 = (e^(i alpha) + 1)`

`=> (z_1 + z_2)/z_1 = e^(i alpha //2) . 2 cos (alpha//2)`

On squaring both sides, we get

`(z_1 + z_2)^2/z_1^2 = e^(i alpha //2) .( 4 cos^2 alpha//2)`

` => (z_1 + z_2)^2/z_1^2 = z_2/z_1 .( 4 cos^2 alpha//2)`

` (z_1 + z_2)^2 = 4 z_1 z_2 cos^2 (alpha//2)`

` (- p)^2 = 4 q cos^2 (alpha //2)`

or `p^2 = 4 q cos^2 (alpha//2)`

Correct Answer is `=>` (D) `4 q cos^2 (alpha// 2)`

Q 2583667547

If `alpha_0 ,alpha_1 , alpha_2 , ... ,alpha_(n-1)` are the n,nth roots of the unity, then find the value of `sum_(i = 0)^(n - 1) (alpha_i)/(2 - alpha _i)`.

(A)

` n/(2^n - 1)`

(B)

` n/(2^n + 1)`

(C)

` 2n/(2^n - 1)`

(D)

None of these

Solution:

Let ` x = (1)^(1//n) => x^n = 1`

`:. x^n - 1 = 0`

or ` x^n - 1 = (x - alpha_0) (x - alpha_1) (x - alpha_2) ... (x - alpha_(n - 1))`

`= prod_(i = 0)^(n -1) (x - alpha_i)`

On taking logarithm both sides, we get

`log_e (x^n -1) = sum_(i = 0)^(n - 1) log_e (x- alpha_i)`

On differentiating both sides w.r.t. x, we get

` (nx^(n-1))/(x^n - 1) = sum_( i = 0)^(n - 1) (1/(x - alpha_i))`

On putting x = 2, we get

` (n(2)^(n -1))/(2^n -1) = sum_(i = 0)^(n - 1) 1/(2 - alpha_i)` .......(i)

Now, `sum_(i = 0)^(n- 1) alpha_i/(2 - alpha_i) = sum_(i = 0)^(n - 1) (- 1 + 2/(2 - alpha _i)) = sum_( i = 0)^( n - 1) 1 + 2 sum_( i = 0)^(n - 1) 1/(2-alpha_i)`

` = - (n) + (2 .n . 2^(n -1) )/(2^n -1)`

` = - n + (n . 2^n)/(2^n - 1) = n/(2^n - 1)`

Q 1483112047

If `alpha, beta, gamma` be the roots of the equation `z^3 + 2z^2 + 2z + 1 = 0` then find the value of
` 57- 50 (alpha^3 + beta^3l + gamma^3 )`.

Solution:

`z^3 + 1 + 2z^2 + 2z = 0`

`=> (z + 1) (z^2- z + 1 + 2z) = 0`

`z = -1, omega, omega^2`

`:.` `57-50 (alpha^3 + beta^3 + gamma^3)`

` =57- 50((-1)^3 + omega^3 + omega^6) = 7`.

Correct Answer is `=>` `7`

Q 1126178071

If `| z - i | ≤ 2` and `z_0 = 5 + 3i`, then the maximum value of `| i z + z_0 |` is

(A)

`10`

(B)

`2−sqrt31`

(C)

`- 14`

(D)

`7`

Solution:

`| i z + z_0 | = | i z - i^2 + z_0 - 1 | = | i ( z - i ) + 5 + 3 i - 1|`

`= | i ( z - i ) + ( 4 = 3i ) |`

`∴ | i z + z_0 | ≤ | i ( z - 2 ) + ( 4 + 3 i ) | ≤ 1.2 + 5 ≤ 7.`

∴ maximum value of ` | i z + z_0 | = 7`

Correct Answer is `=>` (D) `7`

Q 1446491373

If `alpha` is a complex constant such that `alpha z^2 + z +bar alpha= 0` has a
real root, then

(This question may have multiple correct answers)

(A) `alpha+bar alpha=1`

(B) `alpha+ bar alpha=0`

(D) the absolute value of the real root is `1`

Solution:

`alpha z^2 + z + alpha = 0`

Let `z = a` be a real root,

then `alpha a^2 + a+ bar alpha = 0`

and let `alpha = p + iq`

`:. (p + iq) a^2 + a + p - iq = 0`

`=> pa^2 +a+ p = 0` and `a^2q - q = 0`

`:. a=± 1`

`:.` From (i),` alpha ± 1 +alpha= 0`

Also `| a| = 1`.

Correct Answer is `=>` (A)

Q 1452880734

If ` z= (sqrt3-i)/2`, then ` (i^101+z^101)^103 ` equals

(A)

`iz`

(B)

`z`

(C)

`bar(z)`

(D)

None of these

Solution:

`z= (sqrt3-i)/2 =i ((-1-isqrt3/2))= i omega^2`

`:. z ^101 = i^101 omega^202 = i^101 *omega^201*omega = i^101 *omega=i omega`

Now, `i^101 + z^101 = i^101 + i^101 omega = i^101(1+ omega) = i^100*i(-omega^2) = 1 *i(-omega^2) = -i omega^2`

Hence, `(i^101 + z^101)^103 = (-iomega^2)^103 = (-1)^103*i^103*omega^206`

`=(-1)(i^3)(omega^206) = (-1) (-1)(i)(omega^2)=i omega^2=z`

Correct Answer is `=>` (B) `z`

Q 1107323288

Let `z, z_0` be two complex numbers ,`barz_0` being the conjugate of `z_0`. The numbers `z,z_0,zbarz_0` , `1` and ` 0` represented in an Argand diagram by the points `P, P_0,QA` and the origin respectively if `| z | = 1`, then

(This question may have multiple correct answers)

(A) `|z−z_0|=1/2|zbarz_0−1|`

(B) `POP_0` and `AOQ` are congurent

(D) None of these

Solution:

Given `OA = 1, OP = | z | = 1`

OA = OP

`OP_0=|z_0|,OQ=|zbarz_0|=|z_0|=|z_0| `

`OP_0 = OQ` and

`∠POP_0=arg⁡z_0/z`

`∠AOQ=arg⁡1/(zbarz_0)=−arg⁡zbarz_0=arg⁡z_0/z=∠POP_0 `

So triangles are congruent

`PP_0=AQ⇒|z−z_0|=|zbarz_0−1|.`

Correct Answer is `=>` (B)

Q 1147112983

If `|(2iz_1−z_1−z_2)/(2iz_1+z_1+z_2)|=|(cos⁡θ+isinθ)/(cos⁡θ−isinθ)|` and `z_1, z_2, z_3` are non - zero complex numbers, then

(This question may have multiple correct answers)

(A) `(z_1+z_2)/z_1` is purely real

(B) None of these

(D) `(z_1+z_2)/(z_1−z_2)` is purely real

Solution:

we have `|(z_1+i((z_1+z_2))/2)/(z_1−i((z_1+z_2))/2)|=|((2z_1)/(z_1+z_2)+i)/((2z_1)/(z_1+z_2)−i)|=1`

`⇒(2z_1)/(z_1+z_2)` is purely real

`⇒(z_1+z_2)/z_1` is purely real `⇒z_1/z_2` is purely real

Correct Answer is `=>` (A)

Q 1117012880

If `α_1, α_2, ………..α_(n-1)` are roots of the equation `z^(n-1) + z^(n-2) + z^(n-3) + … + z + 1 = 0` ,where `n ≥ 3` and `n ∈ N`, then

(This question may have multiple correct answers)

(A) `α_1, α_2, α_3, ………..α_(n-1)` are in GP.

(B) `ω^n, ω^2n` are the roots of given equation, where `ω, ω^2` are complex cube roots of unity

(D) `ω^(1/n),ω^(2n)` are the roots of given equation

Solution:

Since `z^(n-1) + z^(n-2) + z^(n-3) = ... + z + 1 = 0`

`⇒((z−1))/((z−1)) ( z^(n-1) + z^(n-2) + z^(n-3) + ... + z + 1 = 0) ( z ≠ 1 )`

`⇒ z^n = 1` has roots

`1, α_1, α_2, α_3,...........α_(n-1) ∈ z^n = 1`

Such that `α_k=e^((i2kπ)/n)=(e^((2iπ)/n))^k`

`α_1, α_2, α_3, ………..α_(n-1)` are in GP.,

Correct Answer is `=>` (A) `α_1, α_2, α_3, ………..α_(n-1)` are in GP.

Q 1146178073

`z_1 and barz_1` represent adjacent vertices of a regular polygon of `n` sides whose centres is origin and if `(Im(z_1))/(Re(z_1))=sqrt2−1` then `n` is equal to

(A)

`36`

(B)

`10`

(C)

`8`

(D)

`24`

Solution:

From the figure

`(z_1−0)/(barz_1−0)=(|z_1−0|)/(|barz_1−0|) e^(2lπ//n)=e^(2πi//n)`

`⇒z_1/(barz_1)=e^(2πi//n)` ...............(i)

Let `z^1 = re^(iθ)`, then

`∴ z_1/barz_1=(re^(iθ))/(re^(−iθ))=e^(2iθ)=e^(2πi//n) ` [from(i)]

`∴θ=π/n`

Now

`(Im(z_1))/(Re(z_1))=sqrt2−1 ⇒(rsin(π//n))/(rcos(π/n))=sqrt2−1`

`⇒tan⁡ π/n=tan⁡ π/8`

`∴ n = 8`

Correct Answer is `=>` (C) `8`

Q 2630567412

If `|z| = min (|z- 1 | , | z + 1| }`, then find the value of `|z + bar z|`.

Solution:

If `|z| = |z - 1| => |z|^2 = |z - 1|^2`

`=> z barz = (z- 1) ( bar z - 1) => z + bar z = 1`

again if `|z| = |z + 1|`

` => |z|^2 = |z + 1 => z + bar z = - 1 => | z + bar z | = 1 `Ans.

Q 2650656514

If `| z_2 + iz_1 | = | z_1| + | z_2|` and `lz_1| = 3` & `|z_2| = 4` then area of` Delta ABC`, if affix of A, B & C are `(z_1 ), (z_2)` and `((z_2 - iz_1)/(1 - i) )` respectively, is

(A)

`5/2`

(B)

`0`

(C)

`(25)/2`

(D)

`(25)/4`

Solution:

`|z_2 + iz_1| = | z_1| + |z_2| => Arg (iz_1) + Arg (z_2)`

`=> Arg (z_2) - Arg (z_1) = pi/2`

Let ` z_3 = (z_2 - iz_1)/(1 - i)`

`(1 - i) z_3 = z_2 - iz_1`

`=> (z_3 - z_2) = i (z_3 - z_1)`

`=> (z_2 - z_3) = i (z_1 - z_3)`

`:. angle ABC = pi/2 ` and ` | z_2 - z_3 | = | z_1 - z_3| => AC = BC`

` ∵ AB^2 = AC^2 + BC^2 => AC = 5/sqrt2 ( ∵ AB = 5)`

` :. Delta ABC= 1/2 AC . BC = (AC^2)/2 = (25)/4` square unit

Correct Answer is `=>` (D) `(25)/4`

Q 2660167015

If `z_1 = a+ ib` &` z_2 = c + id` are complex numbers such that `| z_1 | = | z_2 | = 1` and `Re (z_1 z_2) = 0` then the pair of complex numbers `w_1 = a + ic` & `w_2 = b + id` satisfies :

(A)

` | w_1 | = 1 `

(B)

` | w_2 = 1|`

(C)

`Re (w_1 w_2) = 0`

(D)

none

Solution:

`z_1 = a + ib, z_1 = cos theta + i sin theta `

`z_2 = c + id z_2 = cos alpha + i sin alpha`

`z_1z_2 = cos (theta + alpha) + i sin (theta + alpha)`

`:. Re (z_1z_2) = cos (theta + alpha) = 0`

`theta + alpha = pi/2` ..........(i)

`w_1 = cos theta + i cos alpha = cos theta + i cos (pi/2 - 0 )`

` = cos theta + i sin theta = e^(i theta)`

`:. | w_1| =1 `

`w_2 = sin theta + i sin alpha =sin (pi/2 - alpha ) + i sin alpha = cos alpha + i sin alpha = e^(i alpha)`

`:. |w_2| = 1`

`w_1 w_2 = e^(i (theta + alpha))`

`:. Re (w_1 w_2) = cos ( theta + alpha) = 0 ` there ` theta + alpha = pi/2`

Correct Answer is `=>` (A)

Q 2610167019

If `z_1 = 5 + 12 i ` and `|z_2| = 4` then

(A)

maximum `(|z_1 + iz_2 |) = 17`

(B)

minimum `(|z_1 + (1 + i)z_2 |) = 13 - 9 sqrt2`

(C)

minimum ` | z_1/( z_2 + 4/z_2) | = (13)/4`

(D)

minimum ` | z_1/( z_2 + 4/z_2) | = (13)/3`

Solution:

`z_1 = 5 + 12i , |z_2| = 4`

`z_1 + iz_2 | <= |z_1| + |z_2| = 13 + 4 = 17`

`:. | z_1 + (1 + i) z_2 | >= | | z _1| - | 1 + i | z_2 ||`

`= 13 - 4 sqrt2`

`:. min (|z_1 + (1 + i) z_2|) = 13 - 4sqrt2`

` | z_2 + 4/z_2 | <= | z_2 | + 4/(| z_2 |) = 4 + 1 = 5`

` | z_2 + 4/z_2 | >= | z_2 | - 4/(| z_2 |) = 4 - 1 = 3`

`:. max | z_1/( z_2 + 4/z_2) | = (13)/3 ` and ` min | z_1/( z_2 + 4/z_2) | = (13)/5`

Correct Answer is `=>` (A)

Q 2680267117

If `z_1` lies on `|z | = 1` and `z_2` lies on `|z| = 2`, then

(A)

`3 <= |z_1 - 2 z_2| <= 5`

(B)

`1 <= |z_1 + z_2| <= 3`

(C)

`|z_1 - 3z_2| >= 5`

(D)

`|z_1 - z_2| >= 1`

Solution:

`|z_1| = 1, | z_2| = 2`

(A) `| |z_1| - 2 | - z_2| <= | z_1 - 2z_2| <= |z_1| + 2 | - z_2|`

`| 1 - 2(2)| <= | z_1 - 2z_2 | <= 1 + 2(2)`

`3 <= | z_1 - 2z_2 | <= 5`

(B) `| | z_1 - | z_2 | | <= | z_1 + z-2 <= |z_1| + |z_2|`

`| 1 - 2| <= | z_1 + z_2 | <= 1 + 2`

`1 <= | z_1 + z_2 | <= 3`

(C)`||z_1 | - 3 |z_2|| <= | z_1 - 3z_2| <= |z_1| + 3 | - z_2|`

`5 <= | z_1 - 3z_2 | <= 7`

(D) `|| z_1| z-2 <= | z_1 - z_2 | <= | z| + |z_2|`l

`1 <= |z_1 - z_2| <= 3`

Correct Answer is `=>` (A)

Q 2533145942

Find the square root of `x + sqrt(- x^4 - x^2 - 1)`.

(A)

`± ( sqrt((x^2 + 1 + x)/2) - i sqrt((x^2 + 1 - x)/2) )`

(B)

`± ( sqrt((x^2 - 1 + x)/2) + i sqrt((x^2 + 1 - x)/2) )`

(C)

`± ( sqrt((x^2 + 1 + x)/2) + i sqrt((x^2 + 1 - x)/2) )`

(D)

` ( sqrt((x^2 + 1 + x)/2) + i sqrt((x^2 + 1 - x)/2) )`

Solution:

Let `x + sqrt(- x^4 - x^2 - 1)`.

`= x + i sqrt(x^4 + x^2 +1)`

`| z | = sqrt (x^2 + (x^4 + x^2 + 1)) = sqrt( x^4 + 2x^2 + 1) = sqrt((x^2 + 1)^2)`

`| z | = (x^2 + 1), Re (z) = x, Im (z) = sqrt (x^4 + x^2 + 1) > 0`

` sqrtz = ± ( sqrt((| z| + Re (z) )/2) + i sqrt((| z| - Re (z) )/2))`

` :. sqrt( x + sqrt(-x^4 - x^2 - 1 )) = ± ( sqrt((x^2 + 1 + x)/2) + i sqrt((x^2 + 1 - x)/2) )`

Correct Answer is `=>` (C) `± ( sqrt((x^2 + 1 + x)/2) + i sqrt((x^2 + 1 - x)/2) )`

Q 2610367210

If ` ( (1 + i)/(1 - i) )^n = 2/pi ( sec ^(-1) 1/x + sin^(-1) ) x != 0 , - | <= xx <=|`, then find the number of positive integers less than `20` satisfying above equation.

Solution:

` ( (1 + i)/(1 - i) )^n = 2/pi xx pi/2 = 1` Now `( (1 + i)/(1 - i) )^n = 1`

` => i^n = 1 => n = 4 , 8 , 12 ,16 `

Q 2660367215

Find the number of solutions of the equation `z^2 = bar z`.

(A)

(B)

(C)

(D)

Solution:

Let `z = x + iy`

`=> x^2 - y^2 + 2ix y = x - iy => x^2 - y^2 = x ` and ` 2xy = -y`

From (2),` 2xy = - y => y = 0 , x = (-1)/2`

When `y = 0` from (1), we get `x = 0, 1`

when `x = (-1)/2 => y = pm sqrt3/2`

Hence the solution of the equation are

`z_1 = 0 + 1.0 , z_2 = 1 + 1.0 = 1`

`z_3 = (-1)/2 + sqrt3/2 i ` and ` z_4 = (-1)/2 - sqrt3/2 i`

Hence number of solution is `4` Ans.

Correct Answer is `=>` (D) 4

Q 2600891718

If `z_1, z_2, z_3, z_4` are the roots of the equation `z^4 + z^3 + z^2 + z + 1 = 0` then

Column I		Column II
(A)	`\| sum_(i=1)^4 z_i^4\|` is equal to	(P)	`0 `
(B)	`sum_(i=1)^4 z_i^5` is equal to	(Q)	`4`
(C)	`prod_(i=1)^4 (z_i +2)` is equal to	(R)	`1`
(D)	least value of `[\|z_1 + z_2\|]` is (Where `[]` represents greatest integer function)	(S)	`11`
		(T)	`\|4 (cos (pi/3) + i sin (pi/3))\|`

(A)

`A-> s, quad B-> (q,t), quad C->r, quad D->p`

(B)

`A->p, quad B-> (q,t), quad C->s, quad D->r`

(C)

`A-> p, quad B-> (q,t), quad C->r, quad D->s`

(D)

`A-> r, quad B-> (q,t), quad C->s, quad D->p`

Solution:

The given equation is `(z^5-1)/(z-1)=0` which means that `z_1, z_2, z_3, z_4` are four out of five roots of unity except 1.

(A) `z_1^4 +z_2^4+z_3^4 +z_4^4+1^4=0 => | sum_(i=1)^4 z_i^4|=1`

(B) `z_1^5 +z_2^5 +z_3^5 +z_4^5+1^5=5=> sum_(i=1)^4 z_i^5=4`

(C) `z^4+z^3+z^2+z+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)`

Putting `z =- 2` both the sides and we get `prod_(i=1)^4 (z_i+2)=11`

(D) `|z_1+z_2| =sqrt(2+ 2 cos 144^o)` for minimum `=2 cos 72^o=(sqrt 5-1)/2` whose greatest integer is `0`

Correct Answer is `=>` (D) `A-> r, quad B-> (q,t), quad C->s, quad D->p`

Q 2610567419

If `z_1z_2 in C, z_1^2 + z_2^2 in R, z_1 ( z_1^2 - 3z_2^2) = 2` and `z_2 ( 3z_1^2 - z_2^2) = 11`, then find the value of `z_1^2 + z_2^2`

Solution:

Here

`z_1 ( z_1^2 - 3z_2^2 ) = 2` ..... ( 1)

`z_2 ( 3z_1^2 - z_2^2 ) = 11` ..... (2)

multiplying (2) by i and adding it to (1) we get

`z_1^3 - 3z_2^2 z_1 + i (3z_1^2 - z_2^3 ) = 2 + 11 i`

`=> (z_1 + iz_2)^3 = 2 + 11 i => (z_1 - iz_2)^3 = 2 - 11 i`

`=> (z_1^2 + z_2^2)^3 = 125 => z_1^2 + z_2^2 = 5` Ans.