Question

The number of solution(s) of the equation $z^2+|z|=0,$ where $z\in C$ is

Answer 1

Let $z=x+iy$
$z^2+|z|=0$
$\Rightarrow x^2-y^2+2xyi+\sqrt{x^2+y^2}=0$
Comparing real and imaginary parts, we get
$\Rightarrow x^2-y^2+\sqrt{x^2+y^2}=0$
And $2xy=0\Rightarrow x=0$ or $y=0$

$\text{Case}-1:$ When $x=0$
$\Rightarrow -y^2+\sqrt{y^2}=0\Rightarrow y^2=|y|\Rightarrow y=0,\pm 1$
$\therefore$ The solutions are $(0,0)(0,1),(0,-1)$

$\text{Case}-2:$ When $y=0$
$\Rightarrow x^2+\sqrt{x^2}=0\Rightarrow x^2+|x|=0\Rightarrow x=0$
$\therefore$ The solution is $(0,0)$
Hence, the solutions of the equation are $(0,0),(0,1),(0,-1)$