how to calculate the cube root of this complex number [duplicate]

Solution 1:

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?

Solution 2:

$$\large2-2i=\sqrt{8}e^{(-\frac{i\pi}{4}+2ki\pi)}$$.

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.