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Complex number given:

$x = 1 + \cos \alpha + i \sin \alpha$

Desired form is something like $|x| \cdot e^{i \cdot \phi} = |x| \cdot (\cos \phi + i \sin \phi)$.

I somehow got completly stuck how to convert the number to the Euler style.

Maybe someone can help me.

I think I could write:

$x = \cos 0 + i\sin 0 + \cos \alpha + i \sin \alpha$

$x = (\cos 0 + \cos \alpha) + i (\sin 0 + \sin \alpha)$

Then $|x| = \sqrt{(\cos 0 + \cos \alpha)^2 + (\sin 0 + \sin \alpha)^2}$.

Is it then right to write $x = |x| \cdot e^{i \cdot \alpha} = \sqrt{(\cos 0 + \cos \alpha)^2 + (\sin 0 + \sin \alpha)^2} \cdot e^{i \cdot \alpha}$ ?

Is there a simpler way for the Euler style of $x$?

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2 Answers 2

up vote 1 down vote accepted

The last step in your computation is quite wrong as well:

How is $\cos \alpha + \cos 0 + i ( \sin \alpha + \sin 0)$ equal to whatever you have written? Recall, $$\cos A + i\sin A = e^{iA}$$ is Euler's formula and not what you have just written.

This is a standard exercise, so here's the hint:

$$1+ \cos \alpha = 2 \cos^2\frac{\alpha}{2}$$ $$\sin \alpha = 2 \sin \frac {\alpha}{2} \cos \frac {\alpha}{2} $$

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How do i get to these two equations? If I use them now, I can go on, but how do I get $1+cos \alpha = ...$? –  meinzlein Dec 20 '11 at 21:23
    
Recall that $cos \alpha = \cos^2 \frac {\alpha}{2}-\sin^2\frac {\alpha}{2}$ –  user21436 Dec 20 '11 at 21:27
    
Hm this is new to me. We should only use angle sum and difference identities and $sin^2 + cos^2 = 1$ –  meinzlein Dec 20 '11 at 21:36
1  
In what way is this different? You know that $\cos (A+B)=\cos A~\cos B-\sin A \sin B$. Now put $A=B=\frac {\alpha} {2}$ and see! –  user21436 Dec 20 '11 at 21:40

$$ \begin{align} x & = 1 + \cos(\alpha) + i \sin(\alpha) = 2 \cos^2(\alpha/2) + i (2 \sin(\alpha/2) \cos(\alpha/2))\\ & = 2 \cos(\alpha/2) (\cos(\alpha/2) + i \sin(\alpha/2)) = 2 \cos(\alpha/2) e^{i \alpha/2} \end{align}$$

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