Question

Let \(S=\left\{z∈C : \bar{z}=i(z^2+Re(\bar{z}))\right\}\). Then \(\sum\limits_{z∈S}|z|^2\) is equal to 

• A. \(\frac{5}{2}\)
• B. 3
• C. \(\frac{7}{2}\)
• D. 4

Answer 1

The Correct Option is D

Let \( z = x + iy \), where \( x, y \in \mathbb{R} \). Then, \( \overline{z} = x - iy \). Given the equation: \[ \overline{z} = i(z^2 + \operatorname{Re}(\overline{z})) \] Substitute \( z \) and \( \overline{z} \): \[ x - iy = i((x + iy)^2 + x) \] Expand \( z^2 \): \[ (x + iy)^2 = x^2 + 2ixy - y^2 \] Substitute back: \[ x - iy = i(x^2 + 2ixy - y^2 + x) \] Simplify: \[ x - iy = i(x^2 + x - y^2) - 2xy \] Equate real and imaginary parts: \[ x = -2xy \quad \text{and} \quad -y = x^2 + x - y^2 \] Solve \( x = -2xy \): \[ x(1 + 2y) = 0 \implies x = 0 \quad \text{or} \quad y = -\frac{1}{2} \] 
Case 1: \( x = 0 \) Substitute into the imaginary part: \[ -y = 0 + 0 - y^2 \implies y^2 - y = 0 \implies y(y - 1) = 0 \] So, \( y = 0 \) or \( y = 1 \). Thus, \( z = 0 \) or \( z = i \). 
Case 2: \( y = -\frac{1}{2} \) Substitute into the imaginary part: \[ -\left(-\frac{1}{2}\right) = x^2 + x - \left(-\frac{1}{2}\right)^2 \implies \frac{1}{2} = x^2 + x - \frac{1}{4} \] Solve the quadratic equation: \[ x^2 + x - \frac{3}{4} = 0 \implies x = \frac{-1 \pm \sqrt{1 + 3}}{2} = \frac{-1 \pm 2}{2} \] So, \( x = \frac{1}{2} \) or \( x = -\frac{3}{2} \). Thus, \( z = \frac{1}{2} - \frac{1}{2}i \) or \( z = -\frac{3}{2} - \frac{1}{2}i \). The set \( S \) is: \[ S = \{ 0, i, \frac{1}{2} - \frac{1}{2}i, -\frac{3}{2} - \frac{1}{2}i \} \] Calculate \( |z|^2 \) for each \( z \in S \): \[ |0|^2 = 0, \quad |i|^2 = 1, \quad \left| \frac{1}{2} - \frac{1}{2}i \right|^2 = \frac{1}{2}, \quad \left| -\frac{3}{2} - \frac{1}{2}i \right|^2 = \frac{5}{2} \] Sum of \( |z|^2 \): \[ 0 + 1 + \frac{1}{2} + \frac{5}{2} = 4 \] Therefore, the sum \( \sum_{z \in S} |z|^2 \) is equal to \(\boxed{4}\).