If z_1 and z_2 are two complex numbers satisf | vedclass

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(D) Given: $\left|\frac{z_1+z_2}{z_1-z_2}ight|=1$Dividing the numerator and denominator by $z_2$ (assuming $z_2 
eq 0$), we get:$\left|\frac{z_1/z_

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purely imaginary

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(D) Given: $\left|\frac{z_1+z_2}{z_1-z_2}ight|=1$Dividing the numerator and denominator by $z_2$ (assuming $z_2 
eq 0$), we get:$\left|\frac{z_1/z_2 + 1}{z_1/z_2 - 1}ight| = 1$Let $w = \frac{z_1}{z_2}$. Then $|w+1| = |w-1|$.This equation represents the locus of points $w$ that are equidistant from $-1$ and $1$ in the complex plane.The set of points equidistant from two points is the perpendicular bisector of the line segment joining them.The points are $(-1, 0)$ and $(1, 0)$ on the real axis. The perpendicular bisector of the segment joining these points is the imaginary axis.Therefore, $w = \frac{z_1}{z_2}$ must be purely imaginary (i.e.,its real part is $0$).Thus, the correct option is $D$.

If $z_1$ and $z_2$ are two complex numbers satisfying the equation $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$, then $\frac{z_1}{z_2}$ may be

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