Writing the complex number $z = 1 - \sin{\alpha} + i\cos{\alpha}$ in trigonometric form Now I can't finish this problem:
Express the complex number $z = 1 - \sin{\alpha} + i\cos{\alpha}$ in trigonometric form, where $0 < \alpha < \frac{\pi}{2}$.
So the goal is to determine both $r$ and $\theta$ for the expression: $z = r(\cos{\theta} + i\sin{\theta})$
I've done this so far: 


*

*First of all I obtained $r = \sqrt{(1-\sin{\alpha})^2 + \cos^2{\alpha}} = \sqrt{1 + 2 \sin{\alpha} + \sin^2{\alpha} + \cos^2{\alpha}} = \sqrt{2(1 - \sin{\alpha})}$ (possible thanks to the condition over $\alpha$).

*Now I tried to get $\theta = \arctan{\left(\frac{\cos{\alpha}}{1-\sin{\alpha}}\right)}$
And here it is where I get stuck... how to determine $\theta$ with such an expression?
I already know $0 < 1-\sin{\alpha} < 1$ and $0 < \cos{\alpha} < 1$ under the given conditions.
Any help will be appreciated. Thank you :)
P.S. I think (according to my search results here) there are no questions about this problem. I hope you won't mind if it is a duplicate.
 A: A start: To make things more familiar-looking, let $\beta=\frac{\pi}{2}-\alpha$. Then $\sin(\alpha)=\cos(\beta)$ and $\cos(\alpha)=\sin(\beta)$.
Note that by double-angle identities we have $1-\cos(\beta)=2\sin^2(\beta/2)$ and $\sin(\beta)=2\sin(\beta/2)\cos(\beta/2)$.
A: It is worth mentioning that for any $\alpha\in(0,\pi/2)$ we have:
$$\begin{eqnarray*}\frac{\cos\alpha}{1-\sin\alpha}&=&\frac{\sin(\pi/2-\alpha)}{1-\cos(\pi/2-\alpha)}&=&\frac{2\sin(\pi/4-\alpha/2)\cos(\pi/4-\alpha/2)}{2\sin^2(\pi/4-\alpha/2)}\\&=&\cot(\pi/4-\alpha/2)&=&\color{purple}{\tan(\pi/4+\alpha/2)}\end{eqnarray*}$$
hence the argument of your complex number is $\color{purple}{\pi/4+\alpha/2}$ and the square modulus is:
$$\cos^2\alpha+(1-\sin\alpha)^2 = 2-2\sin\alpha=2-2\cos(\pi/2-\alpha)=\color{red}{4\sin^2(\pi/4-\alpha/2)}$$
giving:
$$(1-\sin\alpha)+i\cos\alpha = \color{red}{2\sin(\pi/4-\alpha/2)}\cdot e^{i\color{purple}{(\pi/4+\alpha/2)}}.$$
A: Note that we may express this as $z=1+i(\cos\alpha+i\sin\alpha)=1+i e^{i\alpha}=1+e^{i \alpha+i\pi/2}$ since $i=e^{i\pi/2}$. With a bit of cleverness, we may simplify this immediately: what happens if you pull out a factor of $e^{i\alpha/2+i\pi/4}$?
A: $$z=1-\sin \alpha +i \cos \alpha\\=\{(\cos^2 \dfrac{ \alpha}{2}+\sin^2 \dfrac{ \alpha}{2})-2\sin \dfrac{ \alpha}{2}\cos \dfrac{ \alpha}{2}\}+i(\cos^2 \dfrac{ \alpha}{2}-\sin^2 \dfrac{ \alpha}{2})\\=
(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})^2+i(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2})\\=(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})+i(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2}) \}\\=\sqrt2(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{\cos(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} )+i\sin(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ) \}$$
Hence this complex number has modulus $$r=\sqrt2(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})=2\cos(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} )$$ and the principle argument $$\theta=(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ).$$
