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

Specifically addressing your edit, $\sqrt{20 + 48i} = 2 \sqrt{5 + 12i}$.
Now if $\sqrt{5 + 12i} = a + bi$, then squaring both sides, $5 + 12i = (a^2 - b^2) + 2abi$, or that $5 = a^2 - b^2, 12 = 2ab$. Since $b = \frac{6}{a}$, you can form a quadratic with this information, which gives $a + bi = ±(3 + 2i)$.
