Question

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

• A. $\frac{5}{7}$
• B. $\frac{7}{9}$
• C. $\frac{25}{49}$
• D. none of these

Answer 1

The correct option is D none of these

$\frac{5z_2}{7z_1}=k^1.z^1$

$\frac{z_2}{z_1}=\frac{7k^1}{5}\dot{z}=kz$ [where $k=\frac{7k^1}{5}$]

$\left|\frac{2z_1+3z_2}{2z_1-3z_2}\right|$

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

$=\left|\frac{2+3\times 7k}{2-3\times 7k}\right|$

$=\left|\frac{2+21k}{2-21k}\right|$

$=\frac{|2+21k|}{|2-21k|}$ Both are conjugate pairs so the modulus same.

$=1$