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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:

  1. 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 $\alpha$).

  2. 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.

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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 $\sin(\beta)=2\sin(\beta/2)\cos(\beta/2)$.

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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 $e^{i\alpha/2+i\pi/4}$?

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  • $\begingroup$ I've seen it and it convinces me, but how did you know the factor $e^{i\alpha/2 + i\pi/4}$ would help? Thanks in advance. $\endgroup$
    – user170247
    Aug 21 '14 at 3:26
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    $\begingroup$ Experience, mostly. But the real insight is that that you can write $z$ as $z=1+e^{2i\phi}=e^{i\phi}\left(e^{i\phi}+e^{-i\phi}\right)$ for $\phi=\alpha/2+\pi/4$ which makes the $\cos\phi$ immediately visible. You can also interpret this geometrically: If I have two unit vectors and I add them together, their sum will have the 'average' of their two directions. $\endgroup$ Aug 21 '14 at 3:55
  • $\begingroup$ I learned the geometrical approach at school, but I also wanted to know the analytical one, because nobody knew how to do it :P Anyway, thank you! You've helped me so much! ;) $\endgroup$
    – user170247
    Aug 21 '14 at 5:22
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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: $$(1-\sin\alpha)+i\cos\alpha = \color{red}{2\sin(\pi/4-\alpha/2)}\cdot e^{i\color{purple}{(\pi/4+\alpha/2)}}.$$

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$$z=1-\sin \alpha +i \cos \alpha\\=\{(\cos^2 \dfrac{ \alpha}{2}+\sin^2 \dfrac{ \alpha}{2})-2\sin \dfrac{ \alpha}{2}\cos \dfrac{ \alpha}{2}\}+i(\cos^2 \dfrac{ \alpha}{2}-\sin^2 \dfrac{ \alpha}{2})\\= (\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})^2+i(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2})\\=(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})+i(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2}) \}\\=\sqrt2(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{\cos(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} )+i\sin(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ) \}$$ 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 $$\theta=(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ).$$

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