# How do you find three cube roots of 1+i?

Mar 6, 2017

$\sqrt[6]{2} \left(\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$, $\sqrt[6]{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$ and $\sqrt[6]{2} \left(\cos \left(\frac{17 \pi}{12}\right) + i \sin \left(\frac{17 \pi}{12}\right)\right)$

#### Explanation:

We can use here De Moivre's theorem, which states that if $z = r \left(\cos \theta + i \sin \theta\right)$, then ${z}^{n} = {r}^{n} \left(\cos n \theta + i \sin n \theta\right)$.

It may be worth mentioning that the theorem is valid for fractions as well. As we are going to find cube roots, what we seek is ${\left(1 + i\right)}^{\frac{1}{3}}$.

For that let us first write $1 + i$ in polar form. As $1 + i = 1 + 1 i$, $r = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$ and $\theta = {\tan}^{- 1} \left(\frac{1}{1}\right) = \frac{\pi}{4}$

Hence $1 + i = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

and therefore $\sqrt[3]{1 + i} = {\left(\sqrt{2}\right)}^{\frac{1}{3}} \left(\cos \left(\frac{2 n \pi + \left(\frac{\pi}{4}\right)}{3}\right) + i \sin \left(\frac{2 n \pi + \left(\frac{\pi}{4}\right)}{3}\right)\right)$

Now, choosing n=0,1,2}, we get three roots as

$\sqrt[6]{2} \left(\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$,

$\sqrt[6]{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

and $\sqrt[6]{2} \left(\cos \left(\frac{17 \pi}{12}\right) + i \sin \left(\frac{17 \pi}{12}\right)\right)$