Question

If $z_1$ 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 a positive real part and $z_2$ has negative imaginary part, then $\frac{z_1+z_2}{z_1-z_2}$ 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 $z_1=a+ib=(a,b)$ and $z_2=c-id=(c,d)$,
where $a>0$ and $d>0$
Then, $\left|z_1\right|=\left|z_2\right|$
$\Rightarrow a^2+b^2=c^2+d^2$
Now, $\frac{z_1+z_2}{z_1-z_2}=\frac{(a+ib)+(c-id)}{(a+ib)-(c-id)}$
$=\frac{\left[\right(a+c)+i(b-d\left)\right]}{\left[\right(a-c)+i(b+d\left)\right]}\times \left\{\frac{(a-c)-i(b+d)}{(a-c)-i(b+d)}\right\}$
$=\frac{(a^2+b^2)-(c^2+d^2)-2(ad+bc)i}{a^2+c^2-2ac+b^2+d^2+2bd}$
$=\frac{-(ad+bc)i}{a^2+c^2-ac+bd}$
So, $\frac{z_1+z_2}{z_1-z_2}$ is purely imaginary.
However, if $ad+bc=0$, then $\frac{z_1+z_2}{z_1-z_2}$ will equal to zero.

According to the conditions of the equation, we have $ad+bc=0$.