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# Sir can you tell me the solution of this complex number question in the attachment? The answer is A,D

Vikas TU
14149 Points
4 years ago
put z1 = x1 +iy1
and'z2 = x2 + iy2
simple and also thiss is given as:
(x1)^2 + (y1)^2 = (x2)^2 + (y2)^2.
Us ethis info.
mycroft holmes
272 Points
4 years ago
Several ways to approach this:

1. Write $z = \frac{z_1+z_2}{z_1-z_2} = \frac{1 + \alpha}{1 - \alpha}$ where $\alpha = \frac{z_2}{z_1}$
So

$\overline {z} = \frac{1+\overline {\alpha} } {1-\overline {\alpha} } = \frac{1+\frac{1} {\alpha} } {1- \frac{1} {\alpha}}$

since $|\alpha| =1 \Rightarrow \overline{\alpha} = \frac{1}\alpha$

Hence $\overline{z} = \frac{1+\alpha}{\alpha-1} = -z$

So either z is purely imaginary or z=0. (z=0 is possible when z1=-z2 which is permitted with the restrictions mentioned in the problem)