Show that $\operatorname{Arg}(1+z) = \frac{\theta}{2}$

Question

$\operatorname{Arg}(1+z) = \frac{\theta}{2}$, given that $z = \cos(\theta)+i\sin(\theta)$.

Here is what I have attempted:

• If $z = \operatorname{cis}(\theta)$, then $z+1 = \cos(\theta)+1 +i\sin(\theta)$
• $\tan(\theta) = \frac{y}{x+1} = \frac{\sin(\theta)}{\cos(\theta)+1}$

Is the second step even correct? If so, where do I go from here?

Answer 1

$\frac{\sin(\theta)}{\cos(\theta)+1}=\frac { 2\sin (\theta /2)\cos (\theta /2)} {2\cos^{2}(\theta/2)}=\tan (\theta /2)$.