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

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.)2012-04-13

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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.

enter image description here

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    I used GeoGebra for this one.2012-04-13
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Proof without words:

$\hspace{3.5cm}$proof without words