# How do you convert sqrt(3)i - 1 to polar form?

May 23, 2016

$\left(2 , \frac{2 \pi}{3}\right)$

#### Explanation:

Using the formulae that links Cartesian to Polar coordinates.

•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)

now $\sqrt{3} i - 1 = - 1 + \sqrt{3} i$

here x = -1 and y =$\sqrt{3}$

$\Rightarrow r = \sqrt{{\left(- 1\right)}^{2} + {\left(\sqrt{3}\right)}^{2}} = \sqrt{4} = 2$

and $\theta = {\tan}^{-} 1 \left(- \sqrt{3}\right) = - \frac{\pi}{3}$

$- 1 + \sqrt{3} i \text{ is in the 2nd quadrant}$

so $\theta \text{ requires to be an angle in 2nd quadrant}$

$\Rightarrow \theta = \left(\pi - \frac{\pi}{3}\right) = \frac{2 \pi}{3}$

$\Rightarrow \left(- 1 , \sqrt{3}\right) = \left(2 , \frac{2 \pi}{3}\right)$