Question

If $z_{1}$ and $z_{2}$ are complex numbers such that $z_{1} \neq z_{2}$ and $\left|z_{1}\right|=\left|z_{2}\right|$. If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $\frac{\left(z_{1}+z_{2}\right)}{\left(z_{1}-z_{2}\right)}$ may be

• A. Purely imaginary
• B. Real and positive
• C. Real and negative
• D. None of these

Answer 1

Answer: (A)

Let $z_{1}=a+i b=(a, b)$ and $z_{2}=c-i d=(c,-d)$
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 $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}=\frac{(a+i b)+(c-i d)}{(a+i b)-(c-i d)}$
$=\frac{[(a+c)+i(b-d)][(a-c)-i(b+d)]}{[(a-c)+i(b+d)][(a-c)-i(b+d)]}$
$=\frac{\left(a^{2}+b^{2}\right)-\left(c^{2}+d^{2}\right)-2(a d+b c) i}{a^{2}+c^{2}-2 a c+b^{2}+d^{2}+2 b d}$
$=\frac{-(a d+b c) i}{a^{2}+b^{2}-a c+b d}$ [using (1)]
Hence, $\frac{\left(z_{1}+z_{2}\right)}{\left(z_{1}-z_{2}\right)}$ is purely imaginary.
However if $a d+b c=0$, then $\frac{\left(z_{1}+z_{2}\right)}{\left(z_{1}-z_{2}\right)}$ will be equal to zero.
According to the conditions of the equation, we can have $a d+b c=0$