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
 A: You have$$\frac{1+i}4=\frac1{2\sqrt2}\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)=\frac1{\sqrt2^3}e^{\pi i/4}.$$Can you take it from here?
A: $$\large2-2i=\sqrt{8}e^{(-\frac{i\pi}{4}+2ki\pi)}$$.
So $$\large\frac{1}{2-2i}=\frac{1}{\sqrt{8}}e^{(\frac{i\pi}{4}-2ki\pi)}$$
so $$\Large\frac{1}{(2-2i)^{\frac{1}{3}}}=\frac{1}{\sqrt{8}^{\frac{1}{3}}}e^{\frac{(\frac{i\pi}{4}-2ki\pi)}{3}}=\frac{1}{\sqrt{2}}e^{(\frac{i\pi}{12}-\frac{2ki\pi}{3})}$$.
Where $k$ is an integer.
Now you can write this out in terms of $\cos$ and $\sin$ to get a better representation but this should also suffice.
For $k=0$ you get one cube root (Which is perhaps only what you require as at a beginner level you dont need to be concerned with multiple values).
You get that $$\frac{1}{\sqrt{2}}\left(\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})\right)$$ is one such cube root.
But the expression in the exponential is the general case and it gives you all of the cube roots.
A: 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 $z_1$.
