# Thread: Complex Numbers

1. ## Complex Numbers

Given that z = cos Ѳ + i sin Ѳ show that
2/(1 + z) = 1 – i tan ½ Ѳ

How can I solve that? Please help me.

2. Originally Posted by geton
Given that z = cos Ѳ + i sin Ѳ show that
2/(1 + z) = 1 – i tan ½ Ѳ

How can I solve that? Please help me.
Plug'n'chug:
$\displaystyle \frac{2}{1 + z}$

$\displaystyle = \frac{2}{1 + cos(\theta) + i~sin(\theta)}$

$\displaystyle = \frac{2}{1 + cos(\theta) + i~sin(\theta)} \cdot \frac{1 + cos(\theta) - i~sin(\theta)}{1 + cos(\theta) - i~sin(\theta)}$

$\displaystyle = \frac{2(1 + cos(\theta) - i~sin(\theta))}{(1 + cos(\theta))^2 + sin^2(\theta)}$

$\displaystyle = \frac{2(1 + cos(\theta) - i~sin(\theta))}{1 + 2~cos(\theta) + cos^2(\theta) + sin^2(\theta)}$

$\displaystyle = \frac{2(1 + cos(\theta) - i~sin(\theta))}{2 + 2~cos(\theta)}$

$\displaystyle = \frac{1 + cos(\theta) - i~sin(\theta)}{1 + cos(\theta)}$

$\displaystyle = \frac{1 + cos(\theta)}{1 + cos(\theta)} - i~\frac{sin(\theta)}{1 + cos(\theta)}$

$\displaystyle = 1 - i~\frac{sin(\theta)}{1 + cos(\theta)}$

$\displaystyle = 1 - i~tan \left ( \frac{\theta}{2} \right )$

-Dan