# Why is $|1+e^{i\phi}|=|2\cos(\phi/2)|$?

$$|1+e^{i\phi}|=|2\cos(\phi/2)|$$

Hey guys, just wondering why the above is true, I don't think I quite understand how argand diagrams work. Supposedly, using an argand diagram I should be able to figure that out, but I'm not seeing it.

Ultimately I want to know what $1+ae^{i\phi}$ equates to.

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@mugetsu By what $1+ae^{i \phi}$ equates to, if you mean to ask what complex number is that, or it's modulus, then I have answered that question in essence. (For modulus, just use Euler's form and the definition. You may need $\cos^2 \phi+\sin^2\phi=1$ to simplify.) –  user21436 Apr 13 '12 at 0:41

Euler's formula of complex number gives you that $$e^{ix}=\cos x+ i \sin x$$

The other trigonometric formulas you need here:

$$1+\cos x=2\cos^2\frac x 2\\\sin x=2\sin \frac x 2\cos \frac x 2$$

Here is the computation that uses the formula above:

\begin{align}e^{ix}+1&=1+\cos x+i \sin x\\&=2\cos^2 \frac x 2+i\sin x\\&=2\cos^2\frac x 2+ 2i \sin \frac x 2 \cos \frac x 2 \end{align}

Now, this tells you that $|1+e^{ix}|=|2\cos \frac x 2|$ which is your claim.

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It's pretty natural to view geometrically. Exploit the symmetries of the parallelogram.

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btw, where do you draw such figures? –  Yrogirg Apr 13 '12 at 6:37
I like this answer better. Nice picture and what Yrogirg asked! :-) –  user21436 Apr 13 '12 at 7:36
I used GeoGebra for this one. –  I. J. Kennedy Apr 13 '12 at 14:54
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