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If $Z_1 \neq 0$ and $Z_2$ be two complex numbers such that $\frac{Z_2}{Z_1}$ is a purely imaginary number, then $\left|\frac{2 Z_1+3 Z_2}{2 Z_1-3 Z_2}\right|$ is equal to:
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Let $z_1=1+i$ and $z_2=1-i$
$\frac{z_2}{z_1}=\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=-i$
$\frac{2 z_1+3 z_2}{2 z_1-3 z_2}=\frac{2+3\left(\frac{z_2}{z_1}\right)}{2-3\left(\frac{z_2}{z_1}\right)}=\frac{2-3 i}{2+3 i}$
$\left|\frac{2 z_1+3 z_2}{2 z_1-3 z_2}\right|=\left|\frac{2-3 i}{2+3 i}\right|=\left|\frac{2-3 i}{2+3 i}\right|$
$\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right]$
$=\frac{\sqrt{4+9}}{\sqrt{4+9}}=1$
$\frac{z_2}{z_1}=\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=-i$
$\frac{2 z_1+3 z_2}{2 z_1-3 z_2}=\frac{2+3\left(\frac{z_2}{z_1}\right)}{2-3\left(\frac{z_2}{z_1}\right)}=\frac{2-3 i}{2+3 i}$
$\left|\frac{2 z_1+3 z_2}{2 z_1-3 z_2}\right|=\left|\frac{2-3 i}{2+3 i}\right|=\left|\frac{2-3 i}{2+3 i}\right|$
$\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right]$
$=\frac{\sqrt{4+9}}{\sqrt{4+9}}=1$
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