Question

If $Z_{1} \ne 0$ and $Z_2$ be two complex numbers such that $\frac{Z_{2}}{Z_{1}}$ is a purely imaginary number, then $\left|\frac{2Z_{1}+3Z_{2}}{2Z_{1}-3Z_{2}}\right|$ is equal to:

• A. 2
• B. 5
• C. 3
• D. 1

Answer 1

Answer: (D)

Let $z_{1} = 1 + i$ and $z_{2} = 1 -i$
$\frac{Z_{2}}{z_{1}} = \frac{1-i}{1+i} = \frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)} = -i$
$\frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}} =-\frac{2+3\left(\frac{z_{2}}{z_{1}}\right)}{2-3\left(\frac{z_{2}}{z_{1}}\right)} = \frac{2-3i}{2+3i}$
$\left|\frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}}\right| = \left|\frac{2-3i}{2+3i}\right| = \left|\frac{2-3i}{2+3i}\right|$
$\left[\because \left|\frac{z_{1}}{z_{2}}\right|= \frac{\left|z_{1}\right|}{\left|z_{2}\right|}\right]$
$= \frac{\sqrt{4+9}}{\sqrt{4+9}} = 1$