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: (C)

Re $(z)^{2}=0, |z|=2$
Let $z=x+iy$
$z^{2}=(x+iy)^{2}$
$=x^{2}+i^{2}y^{2}+2ixy$
$z^{2}=(x^{2}-y^{2})+i(2xy)$
$\Rightarrow Re (z^{2})=Re\left\{\left(x^{2}-y^{2}\right)+i\left(2xy\right)\right\}$
$Re \left(z^{2}\right)=x^{2}-y^{2} \ldots\left(i\right)$
$\left|z\right|=\sqrt{x^{2}+y^{2}}=2$
$\Rightarrow x^{2}+y^{2}=4 \ldots\left(ii\right)$
Given, $Re \left(z^{2}\right)=0$
$x^{2}-y^{2}=0 \ldots\left(iii\right)$
On adding Eqs. $\left(ii\right)$ and $\left(iii\right)$
$2x^{2}=4 $
$\Rightarrow x^{2}=2$
$\Rightarrow x=\pm \sqrt{2}$ and $y=\pm \sqrt{2}$
Hence, $z=\pm \sqrt{2}\left(1+i\right)$