Question

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

• A. $3/2$
• B. $1$
• C. $2/3$
• D. $4/9$

Answer 1

Answer: (B)

$1$

$\frac{2z_1}{3z_2}=k^{\prime }i$

$\frac{z_1}{z_2}=\frac{3k^{\prime }i}{2}$

$\frac{z_1}{z_2}=ki$ where $\frac{3k^{\prime }}{2}=k$

$z_1=ki{z}_2$

Thus $\left|\frac{z_1-z_2}{z_1+z_2}\right|=\left|\frac{ki{z}_2-z_2}{ki{z}_2+z_2}\right|$

$=\left|\frac{z_2(ki-1)}{z_2(ki+1)}\right|$

$=\frac{|ki-1|}{|ki+1|}$

$=\frac{\sqrt{k^2+1^2}}{\sqrt{k^2+1^2}}=1$