# How do you multiply: 2(cos(pi/3)+ i sin(pi/3)) and 4(cos(pi/4)+ i sin(pi/4))?

Jun 24, 2016

Product is $8 {e}^{i \frac{7 \pi}{12}} = 8 \left(\cos \left(\frac{7 \pi}{12}\right) + i \sin \left(\frac{7 \pi}{12}\right)\right)$

= $\left(2 \sqrt{2} - 2 \sqrt{6}\right) + i \left(2 \sqrt{6} + 2 \sqrt{2}\right)$

#### Explanation:

To multiply the two complex numbers in polar form there could be three ways.

A - $\left(2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)\right) \times \left(4 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)\right)$

= $8 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)$

= $8 \left(\frac{1}{2 \sqrt{2}} + i \frac{\sqrt{3}}{2 \sqrt{2}} + i \frac{1}{2 \sqrt{2}} - \frac{\sqrt{3}}{2 \sqrt{2}}\right)$

= $\left(2 \sqrt{2} - 2 \sqrt{6}\right) + i \left(2 \sqrt{6} + 2 \sqrt{2}\right)$

B - As $\left(2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)\right) = 2 {e}^{i \left(\frac{\pi}{3}\right)}$and

$4 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) = 4 {e}^{i \left(\frac{\pi}{4}\right)}$, their product is

$4 \times 2 \times {e}^{i \left(\frac{\pi}{3} + \frac{\pi}{4}\right)} = 8 {e}^{i \frac{7 \pi}{12}}$

C - $\left(2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)\right) \times \left(4 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)\right)$

= $8 \left[\left(\cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)\right) + i \left(\sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)\right)\right]$

= $8 \cos \left(\left(\frac{\pi}{3}\right) + \left(\frac{\pi}{4}\right)\right) + i \sin \left(\left(\frac{\pi}{3}\right) + \left(\frac{\pi}{4}\right)\right)$

= $8 \left(\cos \left(\frac{7 \pi}{12}\right) + i \sin \left(\frac{7 \pi}{12}\right)\right)$