Complex solutions of $z^4 = |z|^2 +2$ I'm struggling with the equation in the subject, I managed to find the real solutions ($\pm\sqrt{2}$) by setting $u = z^2$ and I know from Wolfram Alpha that the other two complex solutions should be $\pm i\sqrt{2}$ and I obviouvsly understand why those works but I'm unable to get the correct procedure to find them.
I suspect that there is something around equations with modulus in them that I don't yet know.
Could you point me to the right direction? Thanks!
 A: If $z^4=|z|^2+2$, then$$|z|^4=\left|z^4\right|=\left||z|^2+2\right|=|z|^2+2,$$and therefore $|z|=\pm\sqrt2$.
So, $z=\sqrt2e^{i\theta}$ for some $\theta\in\Bbb R$. And$$z^4=|z|^2+2\iff4e^{4i\theta}=4.$$Can you take it from here?
A: It is possible to find all roots by an easy observation: The right hand side is always real, which means that $z$ either has to be real or purely imaginary so that $z^4\in\mathbb R$.
Let $z=a+bi,$ wehre  $a,b\in\mathbb R$. If you want to find the real solutions, work with $z=a$. Since you were able to find these, I'll skip them.
In order to find the purely imaginary solutions, set $a=0$ so that $z=bi$. Substituting this into our original equation:
$$(bi)^4=\lvert bi\rvert^2+2\Longrightarrow b^4=b^2+2.$$
This is now a simple quadratic equation (set $u=b^2$) you should be able to solve.
A: I believe I managed to solve it by splitting it in two, one with $z>0$ and one with $z<0$ (the latter will give me $z^4 +z^2-2 = 0$)
A: Let $z=x+iy$, where $x,y\in\mathbb{R}$
$z^4=(x+iy)^4=x^4+y^4-6x^2y^2+4ixy(x^2-y^2)$
$|z|^2=x^2+y^2$
$$z^4=|z|^2+2$$
$$\Rightarrow x^4+y^4-6x^2y^2+4ixy(x^2-y^2)=x^2+y^2+2 \ldots (1)$$
Since RHS is real, we must havethe imaginary part of LHS=$0$.
$$\Rightarrow 4xy(x^2-y^2)=0$$
$\Rightarrow x=0$ or $y=0$ or $x^2=y^2$
Using these relations one by one in equation (1), we can get all possible values of $x$ and $y$.
$$\Rightarrow (x,y)=(0,\sqrt2),(0,-\sqrt2),(\sqrt2,0),(-\sqrt2,0)$$
$$\Rightarrow z=\pm\sqrt2,\pm\sqrt2i$$
