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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

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

3 Answers 3

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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?

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  • $\begingroup$ could you please explain the last equality? why do we get $\frac{1}{sqrt{8}}$ * $e^(\frac{pi*i)}{4}$ $\endgroup$
    – user972251
    Dec 22, 2021 at 20:14
  • $\begingroup$ Because $2\sqrt2=\sqrt2^3$ and$$e^{\pi i/4}=\cos\left(\frac\pi4\right)+\sin\left(\frac\pi4\right)i=\frac1{\sqrt2}+\frac i{\sqrt2}.$$ $\endgroup$
    – José Carlos Santos
    Dec 22, 2021 at 23:42
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$$\large2-2i=\sqrt{8}e^{(-\frac{i\pi}{4}+2ki\pi)}$$.

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.

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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 $z_1$.

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