Question

Let $A\left(z_1\right),B\left(z_2\right)$ and $C(z_3)$ 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

Answer: 1.75

$A,B,C$ lie on the unit circle centered at the origin. Also,
$\frac{3z_1+1.0}{4}=\frac{z_2+z_3}{2}$
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 $△BED,$
$\text{BE}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\Rightarrow \left|z_2-z_3\right|=BC=\frac{\sqrt{7}}{2}$
Hence, ${\left|z_2-z_3\right|}^2=\frac{7}{4}=1.75$