# How do I divide 6(cos^circ 60+i\ sin60^circ) by 3(cos^circ 90+i\ sin90^circ)?

Nov 8, 2015

$2 c i s \left(- {30}^{\circ}\right) = \sqrt{3} - i$

#### Explanation:

The easiest way to divide complex numbers is to first convert to polar form and then you may divide modulus separately and subtract the angles.

Let ${z}_{1} = {r}_{1} c i s {\theta}_{1} = 6 c i s {60}^{\circ} = 6 \cos {60}^{\circ} + 6 i \sin {60}^{\circ} = 6 \angle {60}^{\circ} = 6 \angle \frac{\pi}{3}$

${z}_{2} = {r}_{2} c i s {\theta}_{2} = 3 c i s {90}^{\circ} = 3 \cos {90}^{\circ} + 3 i \sin {90}^{\circ} = 3 \angle {90}^{\circ} = 3 \angle \frac{\pi}{2}$

$\therefore {z}_{1} / {z}_{2} = {r}_{1} / {r}_{2} c i s \left({\theta}_{1} - {\theta}_{2}\right)$

$= \frac{6}{3} c i s \left({60}^{\circ} - {90}^{\circ}\right)$

$= 2 c i s \left(- {30}^{\circ}\right) = 2 \angle - {30}^{\circ}$

$= 2 \left(\cos {30}^{\circ} - i \sin {30}^{\circ}\right)$

$= \sqrt{3} - i$