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:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$

There's not realy a need to use De Moivre's Theorem to solve this.
Let a=3-2i:
\begin{align}
   (z-a)^4=z^4 &\Leftrightarrow (z-a)^4 - z^4= 0 \\
& \Leftrightarrow \big((z-a)^2-z^2 \big)\big( (z-a)^2 + z^2\big) =0 \\
& \Leftrightarrow z^2 -2az +a^2 - z^2=0 \quad \vee \quad z^2-2az + a^2 +z^2=0\\
& \Leftrightarrow z=\frac{a}{2} \quad \vee \quad 2z^2 -2az+a^2=0  \\
\end{align}
Solving the quadratic equation
\begin{align}
   2z^2 -2az+a^2=0  & \Leftrightarrow z= \frac{2a \pm \sqrt{4a^2-8a^2}}{4}\\
& \Leftrightarrow z = \frac{2a \pm 2i\sqrt{a^2}}{4} \\
& \Leftrightarrow z=\frac{a \pm ai}{2}
\end{align}
Substituting $a$, we get the following solutions:
$$z=\frac{3}{2}-i \vee z=\frac{5}{2}+\frac{i}{2} \vee z=\frac{1}{2}-\frac{5i}{2}$$
I hope I have been of some assistance.
