# How do you convert 1/sqrt2 - (i)(1/sqrt2) to polar form?

Apr 21, 2018

The polar form is $= \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

#### Explanation:

The polar form of a complex number

$z = a + i b$

is

$z = r \left(\cos \theta + i \sin \theta\right)$

Where,

$r = | z | = \sqrt{{a}^{2} + {b}^{2}}$

$\cos \theta = \frac{a}{| z |}$

and

$\sin \theta = \frac{b}{| z |}$

Here, we have

$z = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$

$| z | = \sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(\frac{1}{\sqrt{2}}\right)}^{2}} = 1$

$\left\{\begin{matrix}\cos \theta = \frac{1}{\sqrt{2}} \\ \sin \theta = - \frac{1}{2}\end{matrix}\right.$

$\iff$, $\theta = - \frac{\pi}{4}$, $\left[\mod 2 \pi\right]$

Therefore,

$z = \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$