If frac{2z_1}{3z_2} is a purely imaginary num | vedclass

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(B) Let $\frac{2z_1}{3z_2} = ki$,where $k \in \mathbb{R}$ and $k 
eq 0$. Then $\frac{z_1}{z_2} = \frac{3}{2}ki$. Let $\frac{z_1}{z_2} = iy$,where $y

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(B) Let $\frac{2z_1}{3z_2} = ki$,where $k \in \mathbb{R}$ and $k 
eq 0$. Then $\frac{z_1}{z_2} = \frac{3}{2}ki$. Let $\frac{z_1}{z_2} = iy$,where $y = \frac{3}{2}k$. Now,consider the expression $\left| \frac{z_1 - z_2}{z_1 + z_2} ight|$. Dividing the numerator and denominator by $z_2$,we get $\left| \frac{\frac{z_1}{z_2} - 1}{\frac{z_1}{z_2} + 1} ight|$. Substituting $\frac{z_1}{z_2} = iy$,we have $\left| \frac{iy - 1}{iy + 1} ight| = \frac{|iy - 1|}{|iy + 1|} = \frac{|-(1 - iy)|}{|1 + iy|} = \frac{|1 - iy|}{|1 + iy|}$. Since the modulus of a complex number $z$ is equal to the modulus of its conjugate $\overline{z}$,and $\overline{1 + iy} = 1 - iy$,we have $|1 - iy| = |1 + iy|$. Therefore,$\frac{|1 - iy|}{|1 + iy|} = 1$.

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

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