# What does (3+i)^(1/3) equal in a+bi form?

Nov 10, 2015

$\sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) i$

#### Explanation:

$3 + i = \sqrt{10} \left(\cos \left(\alpha\right) + i \sin \left(\alpha\right)\right)$ where $\alpha = \arctan \left(\frac{1}{3}\right)$

So

$\sqrt[3]{3 + i} = \sqrt[3]{\sqrt{10}} \left(\cos \left(\frac{\alpha}{3}\right) + i \sin \left(\frac{\alpha}{3}\right)\right)$

$= \sqrt[6]{10} \left(\cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) + i \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right)\right)$

$= \sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) i$

Since $3 + i$ is in Q1, this principal cube root of $3 + i$ is also in Q1.

The two other cube roots of $3 + i$ are expressible using the primitive Complex cube root of unity $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$:

$\omega \left(\sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) i\right)$

$= \sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right) + \frac{2 \pi}{3}\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right) + \frac{2 \pi}{3}\right) i$

${\omega}^{2} \left(\sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right)\right) i\right)$

$= \sqrt[6]{10} \cos \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right) + \frac{4 \pi}{3}\right) + \sqrt[6]{10} \sin \left(\frac{1}{3} \arctan \left(\frac{1}{3}\right) + \frac{4 \pi}{3}\right) i$