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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 $\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} \right|$ is equal to
$2$
$5$
$3$
$1$
Solution
Let ${\text{ }}{z_1} = 1 + i{\text{ }}$ 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 – 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$
