Question

If $z_1=\frac{1}{a+i},a\ne 0$ and $z_2=\frac{1}{1+bi},b\ne 0$ are such that $z_1={\overline{z}}_2$ then

• A. $a=1,b=2$
• B. $a=2,b=1$
• C. $a=-1,b=1$
• D. $a=1,b=-1$

Answer 1

Answer: (D)

$a=1,b=-1$

Given, $z_1=\frac{1}{a+i}=\frac{a-i}{a^2+1}$

$z_2=\frac{1}{1+bi}=\frac{1-bi}{1+b^2}$

Given the condition $z_1=\overline{z_2}$, we have

$\frac{a-i}{a^2+1}=\frac{1+bi}{1+b^2}$
$\Rightarrow \frac{a}{a^2+1}=\frac{1}{1+b^2}$

$\Rightarrow \frac{-i}{a^2+1}=\frac{bi}{1+b^2}$

Dividing the two, we have

$-a=\frac{1}{b}$ or $ab=-1$
Substituting in the first equation, we have $\frac{a}{a^2+1}=\frac{a^2}{1+a^2}$
$\Rightarrow a=0,1$............... $\because a$ cannot be $0$

Thus, $a=1$, implying $b=-1$