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!
 A: Let $w=z+1$ and $w=re^{i\theta}$, then you are given
\begin{align*}
w^3 & = 36\bar{w}\\
w^4 & = 36||w||^2\\
r^2e^{4i \theta} &=36
\end{align*}
Now compare real and imaginary parts.
A: From $(z+1)^3 = 36(z^*+1)$ by applying conjugate we get $(z^*+1)^3 = 36(z+1)$ then $(z^*+1)^9 = 36^3(z+1)^3=36^4(z^* + 1)$. Therefore $z^* + 1 = 0$ or $(z^*+1)^8=36^4$. Should be easy from here. Don't forget the fact that by raising to cube we did introduce some fake solutions, so at the end you'll have to verify all of them. 
A: 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*}
