TITLE: if $a, b$ belong to complex with $|a|=|b|=1$ and $|a+b|=2$ prove that $a=b$
QUESTION [5 upvotes]: I stumbled on this question:

If $a$ and $b$ are complex numbers  with $|a|=|b|=1$ and $|a+b|=2$. Prove that $a=b$. 

I tried to do it using the fact that I can write a and b as  $a= c+zi$ and $b= d+xi$ and when I used the two conditions above I got $|a+b|=2(c^2-d^2) 2cd-2zx =2$, but i don't know how to move on. Could you try to help me?

REPLY [7 votes]: Let $a=\cos\alpha+i\sin\alpha$ and $b=\cos\beta+i\sin\beta$, where $\{\alpha,\beta\}\subset[0,2\pi)$.
Thus,
$$(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=4$$ or
$$\cos(\alpha-\beta)=1$$ or
$$\alpha=\beta.$$
Done!

REPLY [2 votes]: $|a+b|^2=|a|^2+a\bar{b}+b\bar{a}+|b|^2=a\bar{b}+b\bar{a}+2$
Thus $a\bar{b}+b\bar{a}=2$
Note that $a\bar{a}=b\bar{b}=1$
From this we have $\bar{a}=\frac{1}{a}$ and $\bar{b}=\frac{1}{b}$
Thus $\frac{a}{b}+\frac{b}{a}=2\Rightarrow ab(\frac{a}{b}+\frac{b}{a})=2ab \Rightarrow(a-b)^2=0 \Rightarrow a=b$
Also you can prove your result using:
$$a=e^{i\theta_1}$$ $$b=e^{i\theta_2}$$