# How do you convert -1+(sqrt3i) to polar form?

Jan 14, 2017

$\left(r , \theta\right) = \left(2 , {120}^{\circ}\right)$

#### Explanation:

For a complex number in the form
$\textcolor{w h i t e}{\text{XXX}} a + b i$
the radius is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{a}^{2} + {b}^{2}}$
In this case
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(- 1\right)}^{2} _ {\left(\sqrt{3}\right)}^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2$

The angle can be calculated using an inverse trig function and adjusting to place the point in the proper quadrant.

In this case $- 1 + \sqrt{3} i$ is in Quadrant 2 of the complex plane.

Using the $\arctan$ function gives an angle in $\left(- {90}^{\circ} , + {90}^{\circ}\right)$
so it will be necessary to add ${180}^{\circ}$ to shift the angle into Q 2.
$\textcolor{w h i t e}{\text{XXX}} \theta = \arctan \left(- \frac{\sqrt{3}}{1}\right) + {180}^{\circ} = {120}^{\circ}$

(In this particular case, we have one of the standard triangles and the angle could easily be inferred with a quick sketch. )