Question

If $z_{1}and{z}_2$ be complex numbers such that $z_{1}not=z_2$ 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. purely imaginary
• B. Real and positive
• C. Real and negative
• D. None of these

Answer 1

The correct option is A purely imaginary

Let $z_1=a+iband{z}_2=c-id,$ where a $>0$ and d $>$ 0. Then, $\left|z_1\right|=\left|z_2\right|⟹a^2+b^2=c^2+d^2$ Now,
$\frac{z_1+z_2}{z_1-z_2}=\frac{(a+ib)+(c-id)}{(a+ib)-(c-id)}$
= $\frac{[(a+c)+i(b-d)][(a-c)-i(b+d)]}{[(a-c)+i(b+d)][(a-c)-i(b+d)]}$
= $\frac{(a^2+b^2)-(c^2+d^2)-2(ab+bc)i}{a^2+c^2-2ac+b^2+d^2+2bd}$
=$\frac{-(ad+bc)i}{a^2+b^{2}ac+bd}$ [Using (i)
Hence, $(z_1+z_2)/(z_1-z_2)$ is purely imaginary. However, if ad + bc = 0, then $(z_1+z_2)/(z_1-z_2)$ will be equal to zero. According to the conditions of the equation, we can have ad + bc = 0