Complex Numbers of Unit Modulus if $z_1$, $z_2$ and $z_3$ are Complex Numbers of Unit Modulus Such That:
\begin{equation}
|z_1-z_2|^2+|z_1-z_3|^2=4 \tag{1}
\end{equation} Find the value of $$|z_2+z_3|$$
 A: Fix $z_1$ and $z_2$ on the unit circle. The constraint:
\begin{equation}\tag{1} |z_1 - z_2|^2 + |z_1 - z_3|^2 = 4 \end{equation}
Says that $z_3$ must lie on a circle centered at $z_1$ with radius $R = \sqrt{4 - |z_1 - z_2|^2}$. The additional constraint that $z_3$ has unit norm means there are two solutions for $z_3$ which are the intersection points of the two circles (except in the degenerate case when $R = 0$ or $2$ which correspond to $z_2 = \pm z_1$).
Now, the special thing is that this equation is always satisfied when $z_3 = -z_2$. You can see this algebraically, and additionally there is a geometric picture: $z_3 = -z_2$ implies the numbers lie on the same diameter of the unit circle, and so the triangle formed by $z_1,z_2,z_3$ is a right triangle. Then the Pythagorean theorem implies the constraint (1) holds.
From this, you can see that the other solution should be the reflection of $-z_2$ about $z_1$, which is given algebraically by $-z_1^2\bar{z}_2$. In other words: $z_3 = -z_2$ or $z_3 = -z_1^2\bar{z}_2$. In the first case, $|z_2 + z_3| = 0$, and in the second $|z_2 + z_3| = |z_2 - z_1^2\bar{z}_2| = |z_2^2 - z_1^2|$. This can be any value between zero (when $z_2 = \pm z_1$) and two (when $z_2 = \pm i z_1$). So without more specification there is not a unique solution.
A: Tried in this way:
\begin{equation}
|z_1+z_2|^2+|z_1-z_3|^2=4+|z_1+z_2|^2-|z_1-z_2|^2=4+4Re(z_1z_2^*) \tag{2}
\end{equation} $\sim$
\begin{equation}
|z_1+z_3|^2+|z_1-z_2|^2=4+|z_1+z_3|^2-|z_1-z_3|^2=4+4Re(z_1z_3^*) \tag{3}
\end{equation} So
\begin{equation}
|z_2+z_3|^2=|(z_1+z_2)-(z_1-z_3)|^2=\\|z_1+z_2|^2+|z_1-z_3|^2-2Re\left((z_1+z_2)(z_1-z_3)^*\right)=\\4+4Re(z_1z_2^*)-2Re\left((z_1+z_2)(z_1-z_3)^*\right) \tag{4}
\end{equation} 
$\sim$
\begin{equation}
|z_2+z_3|^2=|(z_1+z_3)-(z_1-z_2)|^2=\\|z_1+z_3|^2+|z_1-z_2|^2-2Re\left((z_1+z_3)(z_1-z_2)^*\right)=\\4+4Re(z_1z_3^*)-2Re\left((z_1+z_3)(z_1-z_2)^*\right) \tag{5}
\end{equation} Also Expanding Eqn  $(1)$ We get
\begin{equation}
Re(z_1(z_2+z_3)^*)=0 \tag{6}
\end{equation} Also
\begin{equation}
Re(z-z*)=0 \tag{7}
\end{equation} Adding Egns $(4)$ and $(5)$ and Using $(6)$ and $(7)$ we get
\begin{equation}
|z_2+z_3|^2=2(1+Re(z_2z_3^*))
\end{equation} 
I need further assistance in solving this. I will be happy if there is shorter way to solve preferably using Geometry.
A: The question does not have a well-defined answer. Here are two examples which satisfy the conditions but give different results for $|z_2+z_3|$.
1) $z_2=1$, $z_3=-1$, $|z_1|=1$ (i.e. $z_2$ and $z_3$ are the endpoints of a diameter of the unit circle, and $z_1$ is any point on the circle. By Pythagoras theorem the sum of squares of sides = $(1+1)^2=4$.) Here $|z_2+z_3|=0$.
2) $z_1 = e^{i\pi/2}, z_2 = e^{i\pi/3}, z_3 = e^{-i\pi/3}$. You can verify with WolframAlpha that the condition is satisfied. Here $|z_2+z_3|=2\cos(\pi/3)=1$.
