Question

Let $A\left(z_{1}\right),B\left(z_{2}\right)$ and $C\left(\right.z_{3}\left.\right)$ be complex numbers satisfying the equation $\left|z\right|=1$ and also satisfying the relation $3z_{1}=2z_{2}+2z_{3}$ . Then $\left|z_{2} - z_{3}\right|^{2}$ is equal to

Answer 1

$A,B,C$ lie on the unit circle centered at the origin. Also,
$\frac{3 z_{1} + 1.0}{4}=\frac{z_{2} + z_{3}}{2}$
i.e. the line segment joining $z_{1}$ and origin $\left(D\right)$ bisects the line segment joining $z_{2}\&z_{3}$ at $E$
Also, $DE:EA\equiv 3:1$
Let $DE=3K$ and $EA=K$

Now, $DA=4K=1\Rightarrow K=\frac{1}{4}$
$\Rightarrow DE=\frac{3}{4}$
From $\triangle BED,$
$\textit{BE}=\sqrt{1 - \frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\Rightarrow \left|z_{2} - z_{3}\right|=BC=\frac{\sqrt{7}}{2}$
Hence, $\left|z_{2} - z_{3}\right|^{2}=\frac{7}{4}=1.75$