Question

Let $Z_1,Z_2$ be two complex numbers such that $Z_1\ne Z_2$ and $|Z_1|=|Z_2|$. lf $Z_1$ has positive real part, $Z_2$ has negative imaginary part then $\frac{Z_1+Z_2}{Z_1-Z_2}is$

• A. Complex with non-zero real and imaginary parts
• B. Real and positive
• C. Real and negative
• D. Purely imaginary

Answer 1

Answer: (D)

Purely imaginary

$Z_1,Z_2$ be two complex numbers such that $Z_1{Z}_2$ and $\left|Z_1\right|=\left|Z_2\right|$.
$\frac{(Z_1+Z_2)}{(Z_1-Z_2)}\frac{({\overline{Z}}_1-{\overline{Z}}_2)}{({\overline{Z}}_1-{\overline{Z}}_2)}=\frac{{\left|Z_1\right|}^2-Z_1{\overline{Z}}_2+Z_2{\overline{Z}}_1-{\left|Z_2\right|}^2}{{\left|Z_1\right|}^2-Z_1{\overline{Z}}_2-Z_2{\overline{Z}}_1+{\left|Z_2\right|}^2}$
$=\frac{-(Z_1{\overline{Z}}_2-Z_2{\overline{Z}}_1)}{{\left|Z_1\right|}^2-(Z_1{\overline{Z}}_2+Z_2{\overline{Z}}_1)+{\left|Z_2\right|}^2}$
$=\frac{-2iIm\left(Z_1{\overline{Z}}_2\right)}{{\left|Z_1\right|}^2-2Re\left(Z_1{\overline{Z}}_2\right)+{\left|Z_2\right|}^2}$
Numerator is purely imaginary whereas denominator is purely real.
Note: Im(z) and Re(z) are real numbers.
Hence, $\frac{Z_1+Z_2}{Z_1-Z_2}$ is purely imaginary.