Question

The number of solutions of the system of equations $Re (z^2) = 0, | z | = 2$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

Let $z = x + iy$. Then $z^2 =x^2-y^2 + 2\,i\,xy$
Since $Re (z)^2 = 0$
$\therefore x^{2}-y^{2}=0$
$\therefore y=\pm x$
Now $\left|z\right|=2$
$\Rightarrow x^{2}+y^{2}=4$
$\Rightarrow x^{2}=2$
$\Rightarrow x =\pm\sqrt{2}$
Thus the solutions are
$\left(\sqrt{2}, \sqrt{2}\right), \left(\sqrt{2}, -\sqrt{2}\right), \left(-\sqrt{2}, -\sqrt{2}\right)$,
$\left(-\sqrt{2}, \sqrt{2}\right)$