Question

The complex number(s) $z$ satisfying $Re\left(z^2\right)=0$ and $|z|=\sqrt{3}$ is (are)

• A. $\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i$
• B. $\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i$
• C. $-\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i$
• D. $-\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i$

Answer 1

Answer: (A), (B), (C), (D)

$-\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i$

Let $z=x+iy$
$\Rightarrow z^2=x^2-y^2+2ixy$
$\Rightarrow Re\left(z^2\right)=x^2-y^2=0\cdots \left(1\right)$
Also, $|z|=\sqrt{x^2+y^2}=\sqrt{3}$
$\Rightarrow x^2+y^2=3\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$x=\pm \sqrt{\frac{3}{2}},y=\pm \sqrt{\frac{3}{2}}\Rightarrow z=\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i,\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i,-\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i,-\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i$
Thus, there are four complex numbers.