# How do you express sqrt(2+i) in polar form?

Oct 16, 2016

$\sqrt{2 + i} = {5}^{\frac{1}{4}} {e}^{i \left({\theta}_{0} / 2 + \pi n\right)}$

where $n \in \mathbb{Z}$ and ${\theta}_{0} = \arccos \left(\frac{2}{\sqrt{5}}\right)$

#### Explanation:

The polar form of a complex number is $R {e}^{i \theta}$, where $R$ is the number's modulus (its distance from $0$) and $\theta$ is the angle formed by the positive real axis and the number's vector on the complex plane.

We have a nice way of converting to polar coordinates by using Euler's formula: ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$. Thus, if we can find and factor out $R$, we can find (theta) from the remaining number.

In this case, we will first find $2 + i$ in polar form, and then apply the power of $\frac{1}{2}$.

To find $R$, we find the number's modulus: $| a + b i | = \sqrt{{a}^{2} + {b}^{2}}$

$| 2 + i | = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$

$\implies 2 + i = \sqrt{5} \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}} i\right)$

So, we have $\cos \left(\theta\right) = \frac{2}{\sqrt{5}}$ and $\sin \left(\theta\right) = \frac{1}{\sqrt{5}}$

As $\arccos \left(\frac{2}{\sqrt{5}}\right) = \arcsin \left(\frac{1}{\sqrt{5}}\right) \approx 0.4636$ is not one of the "nice" angles, we'll leave it in that form. For ease of use, let's let ${\theta}_{0} = \arccos \left(\frac{2}{\sqrt{5}}\right)$ and write that for the remainder of the problem.

Proceeding, we now have

$2 + i = \sqrt{5} \left(\cos \left({\theta}_{0}\right) + i \sin \left({\theta}_{0}\right)\right)$

By Euler's formula, this gives us

$2 + i = \sqrt{5} {e}^{i {\theta}_{0}}$

Note that we can add $2 \pi i$ in any integer multiple without changing the value due to the cyclic nature of $\sin \left(\theta\right)$ and $\cos \left(\theta\right)$. This will become relevant once we take the root.

$2 + i = \sqrt{5} {e}^{i \left({\theta}_{0} + 2 \pi n\right)}$

Finally, we take a power of $\frac{1}{2}$ to get

$\sqrt{2 + i} = {\left(\sqrt{5} {e}^{i {\theta}_{0}}\right)}^{\frac{1}{2}}$

$= {5}^{\frac{1}{4}} {e}^{i \frac{{\theta}_{0} + 2 \pi n}{2}}$

$= {5}^{\frac{1}{4}} {e}^{i \left({\theta}_{0} / 2 + \pi n\right)}$