Question

If $z_1$ and $z_2$ are complex numbers such that $z_1\ne z_2$ and $\mid z_1\mid =\mid z_2\mid$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $\left[\right(z_1+z_2\left)/\right(z_1-z_2\left)\right]$ 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 $z_1=a+ib$ and $z_2=c-id$, where $a>0$ and $d>0$. Then,
$|z_1|=|z_2|\Rightarrow a^2+b^2=c^2+d^2$ .....(1)
Now, $\frac{z_1+z_2}{z_1-z_2}=\frac{(a+ib)+(c-id)}{(a+ib)-(c-id)}$
$=\frac{\left[\right(a+c)+i(b-d\left)\right]\left[\right(a-c)-i(b+d\left)\right]}{\left[\right(a-c)+i(b+d\left)\right]\left[\right(a-c)-i(b+d\left)\right]}$
$=\frac{(a^2+b^2)-(c^2+d^2)-2(ad+bc)i}{a^2+c^2-2ac+b^2+d^2+2bd}$
$=\frac{-(ad+bc)i}{a^2+b^2-ac+bd}$ [Using (1)]
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 $ad+bc=0$.