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

Answer: (B)

1

Answer 2

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 $|z|^2$: $|z|^2 = x^2 + y^2 = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$ $|z|^2 = 1$