If $\frac{2 z_1}{3 z_2}$ is a purely imaginary number, then $\left|\frac{z_1-z_2}{z_1+z_2}\right|$ is equal to
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If $\frac{2 z_1}{3 z_2}$ is a purely imaginary number, then $\left|\frac{z_1-z_2}{z_1+z_2}\right|$ is equal to

(A) $3 / 2$

(B) 1

(C) $2 / 3$

(D) $4 / 9$

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SOLUTION —

Let $\frac{2 z_1}{3 z_2}=t y \Rightarrow \frac{z_1}{z_2}=\frac{3}{2} i y$

$\therefore \quad\left|\frac{z_1-z_2}{z_1+z_2}\right|=\frac{\mid \frac{z_1}{z_2}-1}{\mid \frac{z_1}{z_2}+1}\left|=\frac{\mid \frac{3}{2} i y-1}{\mid \frac{3}{2} i y+1}\right|=1 \quad(\because|z|=\mid \bar{z}) \mid$

So, The correct option will be (B).

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