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

Question

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

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.

@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.) — , Apr 13 '12 at 00:41

score 4 · Accepted Answer · 2012-04-13T00:35:04.440

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.

score 4 · Answer 2 · answered Apr 13 '12 at 05:56

4

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

enter image description here

answered Apr 13 '12 at 05:56

I. J. Kennedy

4,048

1

btw, where do you draw such figures? – Yrogirg Apr 13 '12 at 06:37
I like this answer better. Nice picture and what Yrogirg asked! :-) – Apr 13 '12 at 07:36
I used GeoGebra for this one. – I. J. Kennedy Apr 13 '12 at 14:54

score 4 · Answer 3 · answered Apr 14 '12 at 12:10

4

Proof without words:

$\hspace{3.5cm}$ proof without words

answered Apr 14 '12 at 12:10

robjohn

345,667

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

3 Answers3