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Q.
If $\frac{2 z_{1}}{3 z_{2}}$ is a purely imaginary number, then $\left|\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\right|=$
Complex Numbers and Quadratic Equations
Solution:
As given, let $\frac{2 z_{1}}{3 z_{2}}= iy$ or $\frac{z_{1}}{z_{2}}=\frac{3}{2}$ iy, so that
$\left|\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\right|=\left|\frac{\frac{z_{1}}{z_{2}}-1}{\frac{z_{1}}{z_{2}}+1}\right|=\left|\frac{\frac{3}{2} i y-1}{\frac{3}{2} i y+1}\right|=\left|\frac{1-\frac{3}{2} i y}{1+\frac{3}{2} i y}\right|=1 $
$\{\because|z|=|\bar{z}|\}$