Question

Number of solutions of $\mathrm{Re}\left(z^2\right)=0$ and $|z|=a\sqrt{2}$ where $z$ is a complex number and $a>0$ is

• A. $0$
• B. $8$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

$4$

Let, $z=u+iv$

$z^2=u^2-v^2+2iuv$

re(z^2)=0

$u^2=v^2,sou=\pm v$

$given{\left|z\right|}^2=2a^2$

$u^2+v^2=2a^2$

$2u^2=2a^2$

$u=\pm a,v=\pm a$

Therefore, there are 4 solutions.