Argument of $z = 1 - e^{it}$ Let $t\in(0,2\pi)$. How can I find the argument of $z = 1 - e^{it}= 1 - \cos(t) - i\sin(t)$?
 A: Using the inscribed angle theorem I derived $\frac{1}{2}\arg\frac{e^{it}}{-1} = \frac{1}{2}(t - \pi)$ which happens to work for all $t$ in $(0,2\pi)$
A: Since
$$
\begin{align}
\frac{\sin(t)}{1+\cos(t)}
&=\frac{2\sin\left(\frac t2\right)\cos\left(\frac t2\right)}{1+2\cos^2\left(\frac t2\right)-1}\\
&=\frac{\sin\left(\frac t2\right)}{\cos\left(\frac t2\right)}\\[9pt]
&=\tan\left(\tfrac t2\right)\tag{1}
\end{align}
$$
The tangent of the argument of $1-e^{it}$ is
$$
\begin{align}
\frac{-\sin(t)}{1-\cos(t)}
&=-\frac{\sin(\pi-t)}{1+\cos(\pi-t)}\\
&=-\tan\left(\frac{\pi-t}2\right)\\
&=\tan\left(\frac{t-\pi}2\right)\tag{2}
\end{align}
$$
Thus,
$$
\arg(1-e^{it})\equiv\frac{t-\pi}2\pmod\pi\tag{3}
$$
and since the real part of $1-e^{it}$ is $1-\cos(t)\ge0$, we have
$$
-\frac\pi2\le\arg(1-e^{it})\le\frac\pi2\tag{4}
$$
A: Let $t=2s$; multiply and divide by $e^s$:
$$
1-e^{it}=1-e^{2is}=e^{is}(e^{-is}-e^{is})=-e^{is}\cdot2i\sin s
$$
Note that $0<t<2\pi$ implies $0<s<\pi$, so $\sin s>0$. Thus the argument is the same as the argument of $-ie^{is}=e^{i(s-\pi/2)}=e^{i(s+3\pi/2)}$. We choose whichever falls in $[0,2\pi)$. Note that
$$
0\le s-\frac{\pi}{2}<2\pi
$$
means $s\ge \pi/2$, so $\pi\le t<2\pi$, while
$$
0\le s+\frac{3\pi}{2}<2\pi
$$
means $s<\pi/2$, that is $0<t<\pi$.
Therefore the (principal) argument is
$$
\begin{cases}
\dfrac{1}{2}(t+3\pi) & 0<t<\pi\\[12px]
\dfrac{1}{2}(t-\pi)  & \pi\le t<2\pi
\end{cases}
$$
