Question

The number of solutions of the system of equations $Re\left(z^2\right)=0;\left|z\right|=2$ is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is A 4

Let $z=x+iy.$ Then,

$Re\left(z^2\right)=0\Rightarrow Re(x^2-y^2+2ixy)=0$

$\Rightarrow x^2-y^2=0\Rightarrow y=\pm x$ ..... (i)

and, $|z|=2\Rightarrow x^2+y^2=4$ ...... (ii)

Solving (i) and (ii), we get $x=\pm \sqrt{2}$

Thus, the solutions are

$(\sqrt{2},\sqrt{2}),(-\sqrt{2},\sqrt{2}),(\sqrt{2},-\sqrt{2}),(-\sqrt{2},-\sqrt{2})$

Hence, option A.