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

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

• Alt. hint: $\;(1 \pm i)^3 = -2 (1 \mp i)\,$.
– dxiv
Dec 22, 2021 at 20:56

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?

• could you please explain the last equality? why do we get $\frac{1}{sqrt{8}}$ * $e^(\frac{pi*i)}{4}$
– user972251
Dec 22, 2021 at 20:14
• Because $2\sqrt2=\sqrt2^3$ and$$e^{\pi i/4}=\cos\left(\frac\pi4\right)+\sin\left(\frac\pi4\right)i=\frac1{\sqrt2}+\frac i{\sqrt2}.$$ Dec 22, 2021 at 23:42

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

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.

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 $$z_1$$.