Question

If \( z = x + iy \), \( xy \neq 0 \), satisfies the equation \( z^2 + i\overline{z} = 0 \), then \( |z|^2 \) is equal to:

• A. 9
• B. 1
• C. 4
• D. \( \frac{1}{4} \)

Answer 1

The Correct Option is B

Given: \[ z^2 + i\overline{z} = 0 \] where \( z = x + iy \) and \( \overline{z} = x - iy \). 
Substitute \( z = x + iy \): \[ z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy \] and \[ i\overline{z} = i(x - iy) = ix + y \]  
Substitute into the equation: \[ (x^2 - y^2 + 2ixy) + (ix + y) = 0 \] 
Separate the real and imaginary parts: 
From the imaginary part: \[ x(2y + 1) = 0 \] 
Since \( x \neq 0 \), we have \( 2y + 1 = 0 \Rightarrow y = -\frac{1}{2} \).  
Substitute \( y = -\frac{1}{2} \) into the real part: \[ x^2 - \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = 0 \] \[ x^2 - \frac{1}{4} - \frac{1}{2} = 0 \] \[ x^2 = \frac{3}{4} \Rightarrow x = \pm \frac{\sqrt{3}}{2} \]  
Calculate \( |z|^2 \): \[ |z|^2 = x^2 + y^2 = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1 \] \[ |z|^2 = 1 \]