If 2z13z2 is purely, an imaginary number, then z1-z2z1+z2 is equal to

2z13z2=ik ⇒ z1z2=3ik2z1−z2z1+z2=z1/z2−1z1/z2+1=(3ik/2)−1(3ik/2)+1=1

If 2z13z2 is purely, an imaginary number, then z1-z2z1+z2 is equal to

2z13z2=ik ⇒ z1z2=3ik2z1−z2z1+z2=z1/z2−1z1/z2+1=(3ik/2)−1(3ik/2)+1=1

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If $\frac{2 z_{1}}{3 z_{2}}$ is purely, an imaginary number, then $|\frac{z_{1} - z_{2}}{z_{1} + z_{2}}|$ is equal to

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$\frac{2}{3}$

$\frac{3}{2}$

$\frac{4}{9}$

answer is D.

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$\frac{2 z_{1}}{3 z_{2}} = i k \Rightarrow \frac{z_{1}}{z_{2}} = \frac{3 i k}{2}$

$|\frac{z_{1} - z_{2}}{z_{1} + z_{2}}| = |\frac{(z_{1} / z_{2}) - 1}{(z_{1} / z_{2}) + 1}| = |\frac{(3 i k / 2) - 1}{(3 i k / 2) + 1}| = 1$

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If 2z13z2 is purely, an imaginary number, then z1-z2z1+z2 is equal to

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If $\frac{2 z_{1}}{3 z_{2}}$ is purely, an imaginary number, then $|\frac{z_{1} - z_{2}}{z_{1} + z_{2}}|$ is equal to