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 $z=x+iy$
$z^2=(x+iy{)}^2$
$=x^2+i^2{y}^2+2ixy$
$z^2=(x^2-y^2)+i(2xy)$
$\Rightarrow Re(z^2)=Re\left\{\left(x^2-y^2\right)+i\left(2xy\right)\right\}$
$Re\left(z^2\right)=x^2-y^2\dots \left(i\right)$
$\left|z\right|=\sqrt{x^2+y^2}=2$
$\Rightarrow x^2+y^2=4\dots \left(ii\right)$
Given, $Re\left(z^2\right)=0$
$x^2-y^2=0\dots \left(iii\right)$
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, $z=\pm \sqrt{2}\left(1+i\right)$