Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$

To solve
$$
\left(\frac{z-3+2i}{z}\right)^4=1\tag1
$$
Using De Moivre, we want to find $\theta$ so that
$$
\begin{align}
1
&=(\cos(\theta)+i\sin(\theta))^4\tag{2a}\\
&=\cos(4\theta)+i\sin(4\theta)\tag{2b}
\end{align}
$$
which is $\theta\in\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}$
$$
\frac{z-3+2i}z=\cos\left(\tfrac{k\pi}2\right)+i\sin\left(\tfrac{k\pi}2\right)\tag3
$$
for $k\in\{1,2,3\}$ (we won't get a solution for $k=0$). That is,
$$
z=\frac{3-2i}{1-\cos\left(\frac{k\pi}2\right)-i\sin\left(\frac{k\pi}2\right)}\tag4
$$
We get the three solutions
$$
\left\{\frac{5+i}2,\frac{3-2i}2,\frac{1-5i}2\right\}\tag5
$$
