Solve $z^2+|z|=0$ Solve for $z$:
$$z^2+|z|=0$$
I found this question on my textbook but I unable to derive an answer to it as a modulus of $z$ is present. I am unable to take an approach please guide me ..thanks
 A: Let $z=re^{i\theta}$. Then the equation becomes
$$r^2e^{2i\theta}+r=0$$
So either $r=0$ or $re^{2i\theta}=-1$. $$re^{2i\theta}=-1\Leftrightarrow r=1 \text{ and } 2\theta=\pi+2n\pi.$$ 
Plug in we get $z=0$ or $z=i,-i$.
A: Let $z=re^{i\theta}$ then $r^2e^{i2\theta}=-r$ or $re^{i2\theta}=-1$ or  $r=-e^{-i2\theta}$ then $r=1$ and $e^{-i2\theta}=-1$ this concludes
$\cos2\theta-i\sin2\theta=-1$ that is
$$\cos2\theta=-1\,\,\,,\,\,\,\sin2\theta=0$$
thus $2\theta=k\pi$ and $2\theta=2k'\pi+\pi$. you can find $\theta$.
A: put:$z=x+iy$
$$z^2=(x+iy)^2=x^2+i^2y^2+i(2xy)=(x^2-y^2)+i(2xy)\\|z|=\sqrt{x^2+y^2}\\\to z^2+|z|=0\\(x^2-y^2)+i(2xy)+\sqrt{x^2+y^2}=0$$It is better to say $$(x^2-y^2)+i(2xy)+\sqrt{x^2+y^2}=\color{red} {0+0i} \to \begin{cases}(1) \space 2xy=0\\(2) \space (x^2-y^2)+\sqrt{x^2+y^2}= 0\end{cases}\\(1) x=0 \to (2)\space  0-y^2+\sqrt{0+y^2}=0 \to -y^2+|y|=0 \to y=0,1,-1$$
$$(1) \space y=0 \to (2) \space x^2-0+\sqrt{x^2+0}=0 \to \\x^2+|x|=0 \to x=0$$
so 
finally $$(x,y)=(0,0),(0,1),(0,-1)\\z=0+0i \\z=0+i\\z=0-i$$
A: $z^2+|z| = 0$
Clearly $z=0+i\cdot 0$ satisfy above equation.
Now $z^2=-|z|\in \mathbb{R}$
So $z$ must be purely inaginary number.
So Let $z=ki\;, k\in \mathbb{R}$
So put into above equation $-k^2+|k|=0\Rightarrow -|k|^2+|k|=0$
So $-|k|(|k|-1) = 0\Rightarrow |k|=0\Rightarrow k=0$ and $|k| = 1\Rightarrow k=\pm 1$
So we get $z=0,\pm i$ are the solution of $z^2+|z| = 0$ 
Another way::
$z^2+|z|=0\Rightarrow z^2=-|z|\Rightarrow |z|^2=|-|z|| = |z|$
So $|z|^2-|z|=0\Rightarrow |z|(|z|-1)=0\Rightarrow |z|=0\;,|z|=1$
$\bullet\; $ If $|z|=0\;,$ Then put into above equation $z^2+0=0\Rightarrow z=0$
$\bullet\; $ If $|z|=1\;,$ Then put into above equation $z^2+1=0\Rightarrow z=\pm i$
