If $z = (1+\cos a) +i \sin a$ and $a ∈ (0,2π)$, Then find $arg(z)$

Question

This is how I tried to solve ∵ a/2 ∈ (0,π) $$arg(z) =arctan (sin a/ (1+cos a)) =arctan (tan a/2)\\ =a/2 \> when \> a/2 ∈ (0,π/2) \> and \> a/2-π \>when \>a/2 ∈ (π/2,π) $$

Please correct me if I am wrong! Thank you.

Answer 1

Note,

$$z = (1+\cos a) + i \sin a =2\cos^2 \frac a2 +i 2\cos\frac a2\sin \frac a2 = 2\cos\frac a2e^{i\frac a2}$$

For $a\in (0,\pi]$, $arg(z) = \frac a2\in (0,\frac\pi2].$

For $a\in (\pi,2\pi)$, $z= 2|\cos\frac a2|e^{-i\pi+i \frac a2}$ and $arg(z) = -\pi+\frac a2\in (-\frac \pi2,0)$.