Find value of $|z_2+z_3|$ Given that complex numbers $z_1,z_2,z_3$ lie on unit circle and
$$|z_1-z_2|^2+|z_1-z_3|^2=4$$ Then find value of $|z_2+z_3|$
My try:
We can take $z_1=e^{i\alpha}$, $z_2=e^{i\beta}$ and $z_3=e^{i\gamma}$
So we have
$$|z_1-z_2|=2\sin\left(\frac{\alpha-\beta}{2}\right)$$
$$|z_1-z_3|=2\sin\left(\frac{\alpha-\gamma}{2}\right)$$
So we get:
$$\sin^2\left(\frac{\alpha-\beta}{2}\right)+\sin^2\left(\frac{\alpha-\gamma}{2}\right)=1$$
Now we have:
$$|z_2+z_3|=2\cos\left(\frac{\beta-\gamma}{2}\right)$$
Any help here?
 A: As pointed out in comment by Stancek, whom I thank, my previous answer was not correct since it only reported one solution, that was indenpendent on the choice of $z_1$, $z_2$ and $z_3$, i.e 
$$|z_2+z_3|=0,$$
easily found by Pythagorean Theorem.
The general solution can be determined as follows. Use the fact that $|z|^2 = zz^*$, to re-write the equation in the form
$$|z_1|^2 - 2\Re\{z_1z_2^*\} + |z_2|^2 + |z_1|^2 - 2\Re\{z_1z_3^*\} + |z_3|^2=4.$$
Since the points lie on the unit circle, the equation then simplifies to
$$ \Re\{z_1z_2^*\} = -\Re\{z_1z_3^*\}.$$
If you set $z_1 = 1$ you get
$$\Re\{z_2\} = -\Re\{z_3\}.$$
Which yields the two solutions
$$z_3 = -z_2$$
and
$$z_3 = -z_2^*.$$
The first solution leads to 
$$\boxed{|z_2+z_3|=0},$$
that is to $z_1$, $z_2$, and $z_3$ being vertices of a right-angled triangle. 
The second one, in the case $z_1=1$ can be expressed as 
$$|z_2+z_3|= 2\left|\Im\{z_2\}\right|.$$
Taking into account the arbitray phase of $z_1$ yields
$$\boxed{|z_2+z_3|= 2\left|\Im \left\{z_1z_2^*\right\}\right|}.$$
 
A: Note that if $|z|=1$, then $\overline{z}=1/z$. Thus for $z_1$, $z_2$, and $z_3$ on the unit circle, we have
$$\begin{align}
|z_1-z_2|^2+|z_1-z_3|^2
&=\left(z_1-z_2\right)\left({1\over z_1}-{1\over z_2}\right)+\left(z_1-z_3\right)\left({1\over z_1}-{1\over z_3}\right)\\
&=4-\left({z_1\over z_2}+{z_2\over z_1}+{z_1\over z_3}+{z_3\over z_1} \right)\\
&=4-\left((z_1^2+z_2z_3)(z_2+z_3)\over z_1z_2z_3 \right)
\end{align}$$
It follows that $|z_1-z_2|^2+|z_1-z_3|^2=4$ if and only if $(z_1^2+z_2z_3)(z_2+z_3)=0$.  Since we can always arrange for $z_1^2=-z_2z_3$ on the unit circle, $|z_2+z_3|$ can take any value between $0$ and $2$.
A: Hint: Writing $$\frac{\beta-\gamma}{2}=\frac{\beta-\alpha}{2}+\frac{\alpha-\gamma}{2}$$ and combing this with trigonometric identity for the sum of cosine $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$ we get
\begin{align}\cos\left(\frac{\beta-\gamma}{2}\right)&=\cos\left(\frac{\beta-\alpha}{2}+\frac{\alpha-\gamma}{2}\right)\\
&=\cos\left(\frac{\beta-\alpha}{2}\right)\cdot\cos\left(\frac{\alpha-\gamma}{2}\right)-\sin\left(\frac{\beta-\alpha}{2}\right)\cdot\sin\left(\frac{\alpha-\gamma}{2}\right).\end{align}
Now rewrite the identity
$$\sin^2\left(\frac{\alpha-\beta}{2}\right)+\sin^2\left(\frac{\alpha-\gamma}{2}\right)=1$$
as
$$\sin^2\left(\frac{\alpha-\beta}{2}\right)=1-\sin^2\left(\frac{\alpha-\gamma}{2}\right)=\cos^2\left(\frac{\alpha-\gamma}{2}\right),$$
and plug this into the previous to conclude.
