# How do you convert sqrt 2 + sqrt 2i to polar form?

Jul 4, 2016

$\left(2 , \frac{\pi}{4}\right) .$

#### Explanation:

Let $z = \sqrt{2} + \sqrt{2} i = x + i y$, say.

To convert this into polar, we need these Conversion Formula $: x = r \cos \theta , y = r \sin \theta , \theta \in \left(- \pi , \pi\right] , {x}^{2} + {y}^{2} = {r}^{2}$.

We have, $x = \sqrt{2} , y = \sqrt{2}$, so, ${r}^{2} = 2 + 2 = 4 \Rightarrow r = 2$.

Now, $x = r \cos \theta \Rightarrow \sqrt{2} = 2 \cos \theta \Rightarrow \cos \theta = \frac{1}{\sqrt{2}}$.

Similarly, $\sin \theta = \frac{1}{\sqrt{2}}$. Clearly, $\theta = \frac{\pi}{4.}$

Hence, the polar form of the given complex no. is $\left(2 , \frac{\pi}{4}\right) .$