how to calculate the cube root of this complex number Can anyone help me solve this exercise?
calculate the cube roots of $\frac{1}{(2-2i)}$ I started by rationalising by doing $\frac{1}{(2-2i)}$= $\frac{1}{(2-2i)}$ * $\frac{(2+2i)}{(2+2i)}$ --->
$\frac{(2(1+i)}{8}$ ----> $\frac{(1+i)}{4}$
then how can I continue?
thank you all in advance

suppose $z=\dfrac{1+i}{4}=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$. Let $z_1=r(\cos\alpha +\sin\alpha)$ be a root of $z_1^3=z$. Then,
$$r^3(\cos3\alpha+i\sin3\alpha)=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$$
Comparing moduli we get $r^3=\dfrac{1}{2\sqrt{2}}$, hence $r=\dfrac{1}{\sqrt{2}}$. Comparing the stuff inside brackets we get
$$3\alpha=\dfrac{\pi}{4}+2k\pi$$ where $k\in \mathbb{N}$. We get 3 unique values for $\alpha$ which are - $\pi/12$, $3\pi/4$ and $17\pi/12$. So, you have your answer for the three roots by substituting $\alpha$ in $z_1$.
