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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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Verified Answer
The correct answers are:
zero, purely imaginary
Hint : \(\left|\frac{z_1+z_2}{z_1-z_2}\right|=1\)
\(\left|z_1 / z_2+1\right|=\left|z_1 / z_2-1\right|\)
Distance of \(\frac{z_1}{z_2}\) from -1 and 1 are equal
So, locus of \(\frac{z_1}{z_2}\) is perpendicular bisector of line joining \((-1,0) ~\&~(1,0)\)
\(\therefore \frac{\boldsymbol{z}_1}{\mathbf{z}_2} \Rightarrow\) purely imaginary or 0
\(\left|z_1 / z_2+1\right|=\left|z_1 / z_2-1\right|\)
Distance of \(\frac{z_1}{z_2}\) from -1 and 1 are equal
So, locus of \(\frac{z_1}{z_2}\) is perpendicular bisector of line joining \((-1,0) ~\&~(1,0)\)
\(\therefore \frac{\boldsymbol{z}_1}{\mathbf{z}_2} \Rightarrow\) purely imaginary or 0
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