Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1\ne z_2$ and $\left|z_1\right|=\left|z_2\right|$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $(z_1+z_2)/(z_1-z_2)$ may be

• A. zero
• B. real and positive
• C. real and negative
• D. purely imaginary

Answer 1

Answer: (A), (D)

zero
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.