Complex number related problem Let $z_1,z_2,z_3$  be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$
Let ${\theta _1} = 0$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$
$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$
$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$
$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$
$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$
$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$
${\theta _2} \ne {\theta _3}$
How do I proceed further?

This is a kludgy metacheating simplification of the problem.  From the constraints, all 3 points are on the circle of radius 2, centered at the origin.
Further, directly from the constraints, $z_1$ is on the perpendicular bisector of the line segment connecting $z_2$ and $z_3$.  Finally, $z_2, z_3$ must be chosen so that $|z_1 + z_2 + z_3|$ is also on the same circle.  This suggests that the vectors $z_2, z_3$ should cancel each other out, which suggests that $z_2 + z_3$ must equal $0$.
Initially, I considered $(z_2, z_1, z_3) = (-2, 2i, 2)$ as one obvious way of satisfying these constraints.  Then, I realized, based on someone else's (now deleted) comment, that this answer could be harmlessly rotated by any angle $\theta$.
Although I can't prove it, it seems probable to me that the only way to satisfy these constraints is by some rotation of either $(z_2, z_1, z_3) = (-2, 2i, 2)$ or $(z_2, z_1, z_3) = (-2, -2i, 2).$
This means that (for example) the triangle formed by the vertices $(0,0), z_1, z_2$ must be a 45-45-90 right triangle, whose hypotenuse is $|z_1 + z_2| = 2\sqrt{2}$.
Consideration between $z_1$ and $z_3$ is identical.
Therefore, the answer is $\left[2\sqrt{2}\right]^2 = 8.$

Note
A case may be made that this answer is incomplete, since I did not actually prove that there was no other way to satisfy the constraints.  I am therefore relying heavily on the fact that the tone of the question indicates that the answer must be unique.

Addendum
Completing the problem, by proving (for example) that when $z_1 = 2i$ that $z_2, z_3$ must be on the real axis.
I started to analyze this, and then realized that I am merely repeating the analysis already supplied in the answer given by justadzr.
That is, any given $(z_2, z_1, z_3)$ that satisfies the constraints continues to satisfy the constraints when all three points are rotated by $\theta$.  When such a rotation takes $z_1$ to $2i$, then, very similar to what justadzr's answer indicates, $z_2, z_3$ must have form $x + iy, -x +iy$, which subsequently implies that $y = 0$.
This places $z_2, z_3$ on the real axis, which places them at $-2, +2$ (in some order), since these are the only two places where the circle intersects the real axis.
