<s>[INST] Solutions of the complex equation: $(z+1)^3 = 36(z^*+1)$ today in an exam a friend had this equation with complex number:
$$(z+1)^3 = 36(z^*+1)$$
where with z* I mean conjugate z; can someone solve this? 
We tried and there is probably a way to solve it doing a big system with everyterm with i = 0 and everyterm without = 0; but we think that another solution must exists; thanks a lot!
 [/INST] Set $z + 1 = w$, then $\overline{z} + 1 = \overline{w} + 1$ as well, and you want to solve
$$
w^3 = 36\overline{w}.
$$
Multiplying both sides by $w$, we find
$$
w^4 = 36\left|w\right|.
$$
If $w = re^{i\theta}$, then $r^4e^{i4\theta} = 36 r^2$. It follows that either $r = 0$ (in which case $w = 0$), or $r^2 = 36$, so $r = 6$, and that $4\theta = 2k\pi$ for some $k\in\Bbb Z$, or $\theta = k\pi/2$. Since $e^{it}$ is periodic with period $2\pi$, you only need to consider $\theta = 0$, $\theta = \pi/2$, $\theta = \pi$, and $\theta = 3\pi/2$. You get $w = 0$, $w = 6$, $w = 6 i$, $w = -6$, $w = -6i$ as possible solutions. Checking:
\begin{align*}
0^3 &= 36\cdot 0\quad\checkmark\\
6^3&=36\cdot 6\quad\checkmark\\
(6i)^3&=36\cdot (-6i)\quad\checkmark\\
(-6)^3 &=36\cdot(-6)\quad\checkmark\\
(-6i)^3 &=36\cdot(6i)\quad\checkmark\\
\end{align*}
</s>