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. $1$

Answer 1

Answer: (D)

$1$

Solution :

Given that $\frac{5z_2}{11z_1}$ is purely imaginary

Let $\frac{5z_2}{11z_1}=iy$

Now $\frac{z_2}{z_1}=\frac{11iy}{5}$

We have to find $\frac{2z_1+3z_2}{2z_1-3z_2}$

Now,$\frac{2z_1+3z_2}{2z_1-3z_2}=\frac{z_1(2+3\frac{z_2}{z_1})}{z_1(2-\frac{3z_2}{z_1})}=\frac{2+3\times \frac{11iy}{5}}{2-3\times \frac{11iy}{5}}$

$\frac{2z_1+3z_2}{2z_1-3z_2}=\frac{10+33iy}{10-33iy}$

Now $\left|\frac{2z_1+3z_2}{2z_1-3z_2}\right|=\frac{\sqrt{100+1089}}{\sqrt{100+1089}}=1$