A complex Analysis proof Let $a \in \mathbb{C}$ and $\phi \in \mathbb{R}$.
Prove that if $|a+1|=|1+ae^{i \phi}|$ then $ae^{i \phi} = a$ or $ae^{i \phi} = \bar{a}$.
 I  need an idea of how to approach here please anyone.
 A: Think geometrically.  $|a + 1|$ is the distance from $a$ to $-1$, $|1 + a e^{i\phi}|$ is the distance from $a$ to $-e^{-\phi}$.  There are two cases, depending on whether $e^{\phi} = 1$.
A: If $Re(z_1)=Re(z_2)$ and $|z_1|=|z_2|$, then $z_1=z_2$ or $z_1=z_2^*$.
Now $|1+a|=|1+ae^{i\phi}|$ implies $(1+a)(1+a^*)=(1+ae^{i\phi})(1+a^*e^{-i\phi})$.
$$1+a+a^*+|a|^2=1+ae^{i\phi}+a^*e^{-i\phi}+|a|^2$$
$$a+a^*=ae^{i\phi}+a^*e^{-i\phi}$$
$$Re(a)=Re(ae^{i\phi})$$
Besides, $|a|=|ae^{i\phi}|$.
So $ae^{i\phi}=a$ or $a^*$.
A: Let $u = e^{i {\phi \over 2}}$, $\hat{a} = au$, and note that the condition is
equivalent to $|\hat{a} -u| = |\hat{a} - \bar{u}|$.
Since the real parts of
$\hat{a} -u, \hat{a} - \bar{u}$ are the same, Pythagoras gives
$(\operatorname{im}(\hat{a} -u))^2 = (\operatorname{im}(\hat{a} -\bar{u}))^2 $, expanding and simplifying gives $(\operatorname{im}\hat{a}) (\operatorname{im}u) = 0$.
If $\operatorname{im}u = 0$ then $u$ is real and $u^2 =e^{i \phi} =1 $ and so $a= e^{i \phi} a$.
Otherwise $\operatorname{im}\hat{a} = 0$ and so $\hat{a}$ is real. Since $a =  \hat{a} \bar{u}$, we have 
$a e^{i \phi} = a u^2= \hat{a} u = \overline{\hat{a}\bar{u}} = \bar{a}$.
A: $|a+1| = |1+ae^{i \phi}|$  ; $|a+1^2| = |1+ae^{i \phi}|^2$; $(a+1)(\bar a +1) = (1+ae^{i \phi})(1+ \bar{ae^{i \phi}})$; $a \bar a +a+ \bar a + 1 = 1 + ae^{i \phi} + ae^{-i \phi} + a \bar a$ ; $a+ \bar a = ae^{i \phi} + \bar ae^{-i \phi}$; $a-ae^{i \phi} = \bar ae^{-i \phi}- \bar a$; $a(1-e^{i \phi}) = \bar a(e^{-i \phi} -1)$; $a(1-e^{i \phi}) = (\bar a(1-e^{-i \phi}))$ \ $(e^{i \phi})$; $a=\bar a$ \ $e^{i \phi}$. we can now see that $ae^{i \phi} = \bar a$.  
