If {z_1} and {z_2} are two complex numbers sa | vedclass

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(C) Given $\left| \frac{z_1 + z_2}{z_1 - z_2} \right| = 1$. Squaring both sides,we get $\left| \frac{z_1 + z_2}{z_1 - z_2} \right|^2 = 1$. This implie

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Zero or purely imaginary

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(C) Given $\left| \frac{z_1 + z_2}{z_1 - z_2} ight| = 1$. Squaring both sides,we get $\left| \frac{z_1 + z_2}{z_1 - z_2} ight|^2 = 1$. This implies $(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = (z_1 - z_2)(\overline{z_1} - \overline{z_2})$. Expanding both sides: $z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = z_1\overline{z_1} - z_1\overline{z_2} - z_2\overline{z_1} + z_2\overline{z_2}$. Simplifying,we get $2(z_1\overline{z_2} + z_2\overline{z_1}) = 0$,which means $z_1\overline{z_2} + \overline{z_1\overline{z_2}} = 0$. This indicates that $2 	ext{Re}(z_1\overline{z_2}) = 0$,so the real part of $z_1\overline{z_2}$ is $0$. Let $\frac{z_1}{z_2} = x + iy$. Then $z_1 = z_2(x + iy)$. Substituting this into the condition $	ext{Re}(z_1\overline{z_2}) = 0$,we get $	ext{Re}(z_2(x + iy)\overline{z_2}) = 0$,which implies $|z_2|^2 x = 0$. Since $z_2 
eq 0$,we must have $x = 0$. Thus,$\frac{z_1}{z_2}$ is purely imaginary. If $z_1 = 0$,then $\frac{z_1}{z_2} = 0$. Therefore,$\frac{z_1}{z_2}$ is zero or purely imaginary.

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 $\frac{z_1}{z_2}$ is a number which is

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