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?

Algebra + Geometry approach

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_______

Let's try interpret the problem geometrically.
$|z_1| = |z_2 | = |z_3|$ means that the three points lie on a circle.
$|z_1 - z_2 | = |z_1 - z_3|$ means that the chord substended by $z_1$ and $z_2$ is equal to the chord subtended by $z_1$ and $z_3$. This means that $z_1$ is the mid point of the arc on the circle starting at $z_3 $ and ending at $z_2$.
It is also given that $|z_1 + z_2 + z_3 |= 2$, we can rearrange this to give: $|1 + \frac{z_2}{z_1} +\frac{z_3}{z_1} | = 1$ but we know that $\frac{z_2}{z_1} = e^{i \theta}$ and $\frac{z_3}{z_1} = e^{- i \theta}$, this leads to:
$$| 1+ 2 \cos \theta | = 1$$
It is clear the the only two principle solutions for $\theta$ are $\theta = \{\pi, \frac{\pi}{2} \}$... but if $\theta=\pi$ then $z_2 = z_3$ violating the condition in question (draw a picture!)
We go back to the question : $|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|}$
We know when we square a complex number, we square the magnitude and double the angle. Hence, $z_1^2 - z_2^2$ represents a chord having twice the angle as $z_1 - z_2$ and on a circle with squared the radius and similarly does $|z_1 - z_3|^2$.
Ok, but what is the expression for the length of chord which subtends an angle $\theta$ in a circle of radius $r$? By the law of cosines, I find it as $ 2r \sin \frac{\theta}{2}$... now we begin the bash:
$$|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|} =  \frac{ (2r^2 \sin \theta)^2}{(2r \sin \frac{\theta}{2})^2}=4r^2  \cos^2 \frac{\theta}{2}$$
We know $r=2$ and $ \theta= \frac{\pi}{2}$, which gives the final answer as $8$
