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Complex Numbers question

2023 · 13 Apr · Shift 2 · Q24
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Complex Numbers question

2023 · 13 Apr · Shift 2 · Q24

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let S={z∈C:zˉ=i(z2+Re⁡(zˉ))}S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}S={z∈C:zˉ=i(z2+Re(zˉ))}. Then ∑z∈S∣z∣2\sum_{z \in \mathrm{S}}|z|^{2}∑z∈S​∣z∣2 is equal to :
  1. A
    72\frac{7}{2}27​
  2. B
    4
  3. C
    3
  4. D
    52\frac{5}{2}25​
View written solutionFree

Correct answer: B

Let z=x+iyz=x+iyz=x+iy, where x,y∈Rx,y\in\mathbb Rx,y∈R.

We are given zˉ=i(z2+Re⁡(zˉ)).\bar z=i\left(z^2+\operatorname{Re}(\bar z)\right).zˉ=i(z2+Re(zˉ)).

We need to find all such complex numbers zzz, and then compute ∑z∈S∣z∣2.\sum_{z\in S}|z|^2.∑z∈S​∣z∣2.

1. Rewrite all terms in terms of x,yx,yx,y

Since z=x+iy,z=x+iy,z=x+iy, we have zˉ=x−iy,\bar z=x-iy,zˉ=x−iy, and Re⁡(zˉ)=x.\operatorname{Re}(\bar z)=x.Re(zˉ)=x.

Also, z2=(x+iy)2=x2−y2+2ixy.z^2=(x+iy)^2=x^2-y^2+2ixy.z2=(x+iy)2=x2−y2+2ixy.

So, z2+Re⁡(zˉ)=x2−y2+x+2ixy.z^2+\operatorname{Re}(\bar z)=x^2-y^2+x+2ixy.z2+Re(zˉ)=x2−y2+x+2ixy.

Multiplying by iii: i(z2+Re⁡(zˉ))=i(x2−y2+x)+i(2ixy).i\left(z^2+\operatorname{Re}(\bar z)\right)=i(x^2-y^2+x)+i(2ixy).i(z2+Re(zˉ))=i(x2−y2+x)+i(2ixy). Since i2=−1i^2=-1i2=−1, i(2ixy)=−2xy.i(2ixy)=-2xy.i(2ixy)=−2xy. Thus, i(z2+Re⁡(zˉ))=−2xy+i(x2−y2+x).i\left(z^2+\operatorname{Re}(\bar z)\right)=-2xy+i(x^2-y^2+x).i(z2+Re(zˉ))=−2xy+i(x2−y2+x).

Given that zˉ=x−iy=i(z2+Re⁡(zˉ)),\bar z=x-iy=i\left(z^2+\operatorname{Re}(\bar z)\right),zˉ=x−iy=i(z2+Re(zˉ)), we compare real and imaginary parts:

  • Real part: x=−2xyx=-2xyx=−2xy
  • Imaginary part: −y=x2−y2+x-y=x^2-y^2+x−y=x2−y2+x

So the system is x(1+2y)=0x(1+2y)=0x(1+2y)=0 and x2−y2+x+y=0.x^2-y^2+x+y=0.x2−y2+x+y=0.

2. Solve the system

From x(1+2y)=0,x(1+2y)=0,x(1+2y)=0, we get two cases:

Case 1: x=0x=0x=0

Then the second equation becomes 0−y2+0+y=00-y^2+0+y=00−y2+0+y=0 −y2+y=0-y^2+y=0−y2+y=0 y(1−y)=0.y(1-y)=0.y(1−y)=0. So, y=0ory=1.y=0 \quad \text{or} \quad y=1.y=0ory=1.

Thus we get two solutions: z=0,z=i.z=0,\qquad z=i.z=0,z=i.

Case 2: 1+2y=01+2y=01+2y=0

Then y=−12.y=-\frac12.y=−21​. Substitute into x2−y2+x+y=0:x^2-y^2+x+y=0:x2−y2+x+y=0: x2−(14)+x−12=0x^2-\left(\frac14\right)+x-\frac12=0x2−(41​)+x−21​=0 x2+x−34=0.x^2+x-\frac34=0.x2+x−43​=0. Multiply by 444: 4x2+4x−3=0.4x^2+4x-3=0.4x2+4x−3=0. Solve: x=−4±16+488=−4±88.x=\frac{-4\pm\sqrt{16+48}}{8}=\frac{-4\pm 8}{8}.x=8−4±16+48​​=8−4±8​. Hence, x=12orx=−32.x=\frac12 \quad \text{or} \quad x=-\frac32.x=21​orx=−23​.

So two more solutions are z=12−i2,z=−32−i2.z=\frac12-\frac{i}{2},\qquad z=-\frac32-\frac{i}{2}.z=21​−2i​,z=−23​−2i​.

Therefore, S={0, i, 12−i2, −32−i2}.S=\left\{0,\ i,\ \frac12-\frac{i}{2},\ -\frac32-\frac{i}{2}\right\}.S={0, i, 21​−2i​, −23​−2i​}.

3. Compute ∣z∣2|z|^2∣z∣2 for each solution

  1. For z=0z=0z=0: ∣0∣2=0.|0|^2=0.∣0∣2=0.

  2. For z=iz=iz=i: ∣i∣2=1.|i|^2=1.∣i∣2=1.

  3. For z=12−i2z=\frac12-\frac{i}{2}z=21​−2i​: ∣z∣2=(12)2+(−12)2=14+14=12.|z|^2=\left(\frac12\right)^2+\left(-\frac12\right)^2=\frac14+\frac14=\frac12.∣z∣2=(21​)2+(−21​)2=41​+41​=21​.

  4. For z=−32−i2z=-\frac32-\frac{i}{2}z=−23​−2i​: ∣z∣2=(−32)2+(−12)2=94+14=104=52.|z|^2=\left(-\frac32\right)^2+\left(-\frac12\right)^2=\frac94+\frac14=\frac{10}{4}=\frac52.∣z∣2=(−23​)2+(−21​)2=49​+41​=410​=25​.

Now add them: ∑z∈S∣z∣2=0+1+12+52=4.\sum_{z\in S}|z|^2=0+1+\frac12+\frac52=4.∑z∈S​∣z∣2=0+1+21​+25​=4.

4. Final answer

Thus, ∑z∈S∣z∣2=4.\sum_{z\in S}|z|^2=4.∑z∈S​∣z∣2=4.

So the correct option is B.

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