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Question: The complex numbers z<sub>1</sub> and z<sub>2</sub> are such that z<sub>1</sub> ≠ z<sub>2</sub> and ...

The complex numbers z1 and z2 are such that z1 ≠ z2 and z1=z2. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2z1−z2)\left( \frac{z_{1} + z_{2}}{z_{1} - z_{2}} \right)may be

A

Zero

B

Real and positive

C

Real and negative

D

Purely imaginary

Answer

Purely imaginary

Explanation

Solution

Sol. We have z1=z2, Re(z1) > 0, Im( z2) < 0.

Re(z1+z2z1−z2)=12(z1+z2z1−z2+zˉ1+zˉ2zˉ1−zˉ2)\left( \frac{z_{1} + z_{2}}{z_{1} - z_{2}} \right) = \frac{1}{2}\left( \frac{z_{1} + z_{2}}{z_{1} - z_{2}} + \frac{{\bar{z}}_{1} + {\bar{z}}_{2}}{{\bar{z}}_{1} - {\bar{z}}_{2}} \right)

= 12((z1+z2)(zˉ1−zˉ2)+(zˉ1+zˉ2)(z1−z2)(z1−z2)(zˉ1−zˉ2))\frac{1}{2}\left( \frac{\left( z_{1} + z_{2} \right)\left( {\bar{z}}_{1} - {\bar{z}}_{2} \right) + \left( {\bar{z}}_{1} + {\bar{z}}_{2} \right)\left( z_{1} - z_{2} \right)}{\left( z_{1} - z_{2} \right)\left( {\bar{z}}_{1} - {\bar{z}}_{2} \right)} \right)

= 12(z1zˉ1−z1zˉ2+z2zˉ1−z2zˉ2+z1zˉ1+z1zˉ2−z2zˉ1−z2zˉ2∣z1−z2∣2)\frac{1}{2}\left( \frac{z_{1}{\bar{z}}_{1} - z_{1}{\bar{z}}_{2} + z_{2}{\bar{z}}_{1} - z_{2}{\bar{z}}_{2} + z_{1}{\bar{z}}_{1} + z_{1}{\bar{z}}_{2} - z_{2}{\bar{z}}_{1} - z_{2}{\bar{z}}_{2}}{\left| z_{1} - z_{2} \right|^{2}} \right)

=12(2∣z1∣2−2∣z2∣2∣z1−z2∣2)\frac{1}{2}\left( \frac{2\left| z_{1} \right|^{2} - 2\left| z_{2} \right|^{2}}{\left| z_{1} - z_{2} \right|^{2}} \right)= 0 ⇒ (z1+z2z1−z2)\left( \frac{z_{1} + z_{2}}{z_{1} - z_{2}} \right) is purely imaginary