Question

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

• A. $\frac{37}{33}$
• B. $\frac{11}{5}$
• C. $1$
• D. None of these

Answer 1

Answer: (C)

$1$

$\frac{5z_2}{7z_1}=ki(k\in R),$ then $\frac{z_1}{z_2}=\left(\frac{7k}{5}\right)i$

$\Rightarrow$ $\left|\frac{2z_1+3z_2}{2z_1-3z_2}\right|=\left|\frac{2\left(\frac{z_1}{z_2}\right)+3}{2\left(\frac{z_1}{z_2}\right)-3}\right|$

$=\left|\frac{14ki+15}{14ki-15}\right|$

$=\left|\frac{\sqrt{196k^2+225}}{\sqrt{196k^2+225}}\right|$

$=1$