# How do you convert 3-2i  to polar form?

Jun 8, 2018

$\sqrt{13} \setminus \textrm{c i s} \left(\textrm{\arctan} \left(- \frac{2}{3}\right)\right)$

#### Explanation:

Convert $z = 3 - 2 i$ to polar form

$| z | = \setminus \sqrt{{\left(3\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{13}$

$\angle z = \textrm{\arctan 2} \left(- 2 \text{ /, } 3\right)$

I write the angle this way to correct the usual error. Normally people use the single parameter inverse tangent, which only covers two quadrants. We need the two parameter, four quadrant inverse tangent. The way it works is:

$\textrm{\arctan 2} \left(b \text{ /, } a\right) = \arctan \left(\frac{b}{a}\right) \mathmr{if} a > 0$

$\textrm{\arctan 2} \left(b \text{ /, } a\right) = \pi - \arctan \left(\frac{b}{a}\right) \mathmr{if} a < 0$

$\textrm{\arctan 2} \left(b \text{ /, } a\right) = \textrm{s g n} \left(b\right) \frac{\pi}{2} \mathmr{if} a = 0$

We have positive $a$ so first case, fourth quadrant;

angle z = text{arctan}(-2 / 3) approx -33.69°

$z = | z | \textrm{c i s} \left(\angle z\right)$

$z = \sqrt{13} \setminus \textrm{c i s} \left(\textrm{\arctan} \left(- \frac{2}{3}\right)\right)$

where of course $\textrm{c i s} \setminus \theta = \cos \theta + i \sin \theta .$