Find the amplitude of a complex number 
If 
$$z = (1-\cos{\theta}) + i \sin{\theta},$$
then the amplitude of $z$ is:
$$a) \frac{\pi}{4}-\frac{\theta}{2}$$ $$b)
 \frac{\pi}{2}-\frac{\theta}{2}$$ $$c) \frac{\pi}{2}-\theta$$ $$d)
 \frac{\pi}{2}-\frac{\theta}{4}$$ where 0<$\theta$ <$\pi$.

I tried :
$x= 1-\cos{\theta}, y= \sin{\theta}$ then 
$\tan{\phi}=\frac{\sin{\theta}}{1-\cos{\theta}}$.
However I could not solve the last equation.
 A: $\sin\theta = 2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}$
$1-\cos\theta = 2\sin^2\frac{\theta}{2}$
So, $\tan\phi = \frac{\sin\theta}{1-\cos\theta} = \frac{2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$

$\tan\phi = \cot\frac{\theta}{2} = \tan(\frac{\pi}{2} - \frac{\theta}{2})$
$\implies\phi =\frac{\pi}{2} - \frac{\theta}{2} $

For half-angle formulas: https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php
A: Your $z$ can be nicely written as
$$z=1-e^{-i \theta}$$
Now, 
\begin{align}
|z| &= |1-e^{-i \theta}| \\
&=\bigg |e^{-i \frac{\theta}{2}}\left(e^{i \frac{i\theta}{2}} -e^{-i \frac{\theta}{2}}  \right)\bigg | \\
&=\bigg |e^{-i \frac{i\theta}{2}} \left( \cos \left( \frac{\theta}{2}\right) + i \sin  \left( \frac{\theta}{2}\right)-\cos \left( \frac{-\theta}{2}\right) - i \sin  \left( \frac{-\theta}{2}\right) \right)\bigg |\\
&= \bigg |2i \sin \left( \frac{\theta}{2}\right)e^{i \frac{i\theta}{2}}\bigg|\\
&=  |2i|. \bigg |\sin \left( \frac{\theta}{2}\right)\bigg|.\bigg|e^{i \frac{-\theta}{2}}\bigg| \\
&= 2 \bigg |\sin \left(\frac{\theta}{2}\right)\bigg | 
\end{align}
Whence
\begin{align}
\angle z &= \angle (2i)+\angle\left( \sin \left(\frac{\theta}{2}\right) \right)+\angle (e^{-i \frac{\theta}{2}}) \\
&= \frac{\pi}{2}+0-\frac{\theta}{2}
\end{align}
as required.
A: If $z=1-\dfrac{1-t^2}{1+t^2}+i\dfrac{2t}{1+t^2}=\dfrac{2t^2}{1+t^2}+i\dfrac{2t}{1+t^2}$ where $t=\tan\dfrac\theta2$ 
So, if arg$(z)=u,\tan u=\dfrac{\dfrac{2t}{1+t^2}}{\dfrac{2t^2}{1+t^2}}=\dfrac1t=\tan\left(\dfrac\pi2-\dfrac\theta2\right)$
Using this, $$u=\begin{cases}\dfrac\pi2-\dfrac\theta2 &\mbox{if } t>0 \\
\pi+\left(\dfrac\pi2-\dfrac\theta2\right)& \mbox{if } t<0\\\text{undefined}  &\mbox{if }t=0\end{cases}$$
A: Let $z=a+ib$, first find  $\theta_0=\tan^{-1} |b/a|, \theta_0$ must be an acute angle in $[0,\pi/2]$. Then find which quadrant of $x-y$ plane $z$ belongs to.  The principal value of $amp(z)$  lies in $(-\pi, \pi]$. $ amp(z)= \theta_0, \pi-\theta_0, \theta_0-\pi, ~ \mbox{and}~-\theta_0$, if $z$ lies in the I, II, III and IV quadrant respectively.
When $z=(1-\cos \theta)+ i\sin \theta$, it lies only in I or IV quadrant as 
$(1-\cos \theta)=2 \sin^2(\theta/2)>0$. Here $\tan\theta_0=\tan^{-1} |\cot(\theta/2)|=|\pi/2-\theta/2|$. So $amp(z)=\pi/2-\theta/2$, it will be positive if $\theta \in (0, \pi)$ and negative if $\theta \in (\pi, 2\pi)$. 
