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Question: If \(\left| \frac{z_{1}z - z_{2}}{z_{1}z + z_{2}} \right|\)= k, (z<sub>1</sub>, z<sub>2</sub> ≠ 0) t...

If ∣z1z−z2z1z+z2∣\left| \frac{z_{1}z - z_{2}}{z_{1}z + z_{2}} \right|= k, (z1, z2 ≠ 0) then

A

For k = 0 , locus of z is a straight line

B

For k ∉ {1, 0}, z lies on a circle

C

For k = 1, z represents a point

D

For k = 0, z lies on the perpendicular bisector of the line segment joiningz2z1\frac{z_{2}}{z_{1}}and –z2z1\frac{z_{2}}{z_{1}}

Answer

For k ∉ {1, 0}, z lies on a circle

Explanation

Solution

Sol. The expression ∣z1z−z2z1z+z2∣\left| \frac{z_{1}z - z_{2}}{z_{1}z + z_{2}} \right| = k ⇒ ∣z−z2z1z+z2z1∣=k\left| \frac{z - \frac{z_{2}}{z_{1}}}{z + \frac{z_{2}}{z_{1}}} \right| = k

⇒ (z−z2z1)(zˉ−zˉ2zˉ1)=k2(z+z2z1)(zˉ+zˉ2zˉ1)\left( z - \frac{z_{2}}{z_{1}} \right)\left( \bar{z} - \frac{{\bar{z}}_{2}}{{\bar{z}}_{1}} \right) = k^{2}\left( z + \frac{z_{2}}{z_{1}} \right)\left( \bar{z} + \frac{{\bar{z}}_{2}}{{\bar{z}}_{1}} \right)

⇒ ∣z−(1+k2)1−k2.z2z1∣=2k1−k2∣z2z1∣\left| z - \frac{\left( 1 + k^{2} \right)}{1 - k^{2}}.\frac{z_{2}}{z_{1}} \right| = \sqrt{\frac{2k}{1 - k^{2}}}\left| \frac{z_{2}}{z_{1}} \right|

For k ≠ 0, 1, this represents a circle