if $a, b$ belong to complex with $|a|=|b|=1$ and $|a+b|=2$ prove that $a=b$ 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? 
 A: 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!
A: Extended hint. $$\;4 = |a+b|^2 = (a+b)(\bar a + \bar b) = |a|^2+ a \bar b + \bar a b + |b|^2 = 2 + a \bar b + \bar a b \implies a\bar b+ \bar a b - 2= 0$$
But $|a|=1 \iff \bar a = 1/a$ and the same holds for $b$, so $a/b + b/a - 2 = 0 \iff (a-b)^2=0\,$.
A: by the triangle inequality we get $$|a+b|\le |a|+|b|$$ since we have $$|a|=|b|=1$$ we get the equal sign and therefore we get $$a=b$$
A: $|a+b|^2=|a|^2+a\bar{b}+b\bar{a}+|b|^2=a\bar{b}+b\bar{a}+2$
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:
$$a=e^{i\theta_1}$$ $$b=e^{i\theta_2}$$
A: Think about vector form $$|a+b|=\sqrt{|a|^2+|b|^2+2|a||b|\cos \alpha}\\\to 
2=\sqrt{1+1+2.1.1\cos \alpha}\\4=2+2\cos \alpha \\\to\cos \alpha=1 \to \alpha=0 \\so\\|a|=|b|=1 , \alpha=0 \implies a=b$$
