Question

If $\frac{5z_1}{7z_2}$ is purely imaginary, then $\left|\frac{2z_1+3z_2}{2z_1-3z_2}\right|is$

• A. $\frac{5}{7}$
• B. $\frac{7}{5}$
• C. $1$
• D. $\frac{3}{5}$

Answer 1

The correct option is C $1$

$\frac{5}{7}\frac{z_1}{z_2}$ is purely imaginary $\Rightarrow \frac{z_1}{z_2}$ is purely imaginary

$\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=0\Rightarrow z_1\overline{z_2}+z_2\overline{z_1}=0--\left(1\right)$

Now,
${\left|\frac{2z_1+3z_2}{2z_1-3z_2}\right|}^2=\frac{(2z_1+3z_2)\times (2\overline{z_1}+3\overline{z_2})}{(2z_1-3z_2)\times (2\overline{z_1}-3\overline{z_2})}\{as|z{|}^2=z.\overline{z}\}$

$=\frac{4|z_1{|}^2+6z_1\overline{z_2}+6z_2\overline{z_1}+9|z_2^2|}{4|z_1{|}^2-6z_1\overline{z_2}-6z_2\overline{z_1}+9|z_2{|}^2}$

$=\frac{4|z_1{|}^2+9|z_2{|}^2+6(z_1\overline{z_2}+z_2\overline{z_1})}{4|z_1{|}^2+9|z_2{|}^2-6(z_2\overline{z_1}+z_1\overline{z_2})}$

But from (1) ,$z_1\overline{z_2}+z_2\overline{z_1}=0$

$=\frac{4|z_1{|}^2+9|z_2{|}^2}{4|z_1{|}^2+9|z_2{|}^2}=1$