<s>[INST] Find all the roots of the equation $z^2 + |z| = 0$ . Find all the roots of the equation $z^2 + |z| = 0$ .
My solution goes like this:

Let $z=a+ib$, and $z^4=|z|^2\implies a^4+4a^3ib+6a^2(ib)^2+4a(ib)^3+(ib)^4=a^4+4a^3ib-6a^2b^2-4ab^3i=a^2+b^2\implies a^4+b^4+4a^3ib-6a^2b^2-4aib^3+b^4=a^2+b^2\implies a^4+b^4+4i(a^3-b^3)=a^2+b^2+6a^2b^2\implies a^4+b^4-2a^2b^2+4i(a^3-b^3)=a^2+b^2+4a^2b^2\implies (a^2-b^2)^2+4i(a-b)(a^2+b^2+ab)=a^2+b^2+4a^2b^2$. Now, since, $\text{R.H.S} \in\Bbb R$ and hence $Im(\text{L.H.S})=0$. Thus, $4(a-b)(a^2+b^2+ab)=0$ so, $a=b$ and $a^2+b^2+ab$. So all $z=a+ib$ are solutions of the equations iff $a=b$ or $\text{if}\space(x=a,y=b)$ satisfies $f(x,y)=x^2+y^2+xy=0$.

Is the above solution correct?
 [/INST] If $z^2+|z|=0$, then $z^2=-|z|\leqslant0$. The only complex numbers whose square is a real number smaller than or equal to $0$ are those numbers $z$ of the form $\lambda i$, for some $\lambda\in\Bbb R$. But then$$z^2+|z|=0\iff-\lambda^2+|\lambda|=0,$$whose only solutions are $-1$, $0$, and $1$. Therefore, the solutions of your equation are $-i$, $0$, and $i$.
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