Question

Let a and b be two fixed non zero complex numbers and z is a variable complex number. If the lines $a\overline{z}+\overline{a}z+1=0$ and $b\overline{z}+\overline{b}z-1=0$ are mutually perpendicular, then

• A. $ab+\overline{a}\overline{b}=0$
• B. $ab-\overline{a}\overline{b}=0$
• C. $\overline{a}b-a\overline{b}=0$
• D. $a\overline{b}+\overline{a}b=0$

Answer 1

The correct option is D $a\overline{b}+\overline{a}b=0$

Let the equation of straight line be $lx+my+n=0$ ...(1)
where $\frac{-l}{m}$ is slope.
Let $z=x+iy$
$\therefore \overline{z}=x-iy\Rightarrow x=\frac{z+\overline{z}}{2}\&y=\frac{z-\overline{z}}{2i}$
Substitute $x$ & $y$ in equation (1), we get
$l\left(\frac{z+\overline{z}}{2}\right)+m\left(\frac{z-\overline{z}}{2i}\right)+n=0\Rightarrow \frac{1}{2}(l-im)z+\frac{1}{2}(l+im)\overline{z}+n=0$
Let $a=\frac{l+im}{2}\Rightarrow \overline{a}=\frac{l-im}{2}$
$\therefore l=\frac{a+\overline{a}}{2}\&m=\frac{a-\overline{a}}{2i}$
Therefore slope of straight line is $\frac{-l}{m}=\frac{i(a+\overline{a})}{a-\overline{a}}$.
Slope of straight line $a\overline{z}+\overline{a}z+1=0$ is $m_1=\frac{i(a+\overline{a})}{a-\overline{a}}$
Slope of straight line $b\overline{z}+\overline{b}z-1=0$ is $m_2=\frac{i(b+\overline{b})}{b-\overline{b}}$
Since both straight lines are perpendicular.
Therefore $m_1{m}_2=-1$
$\Rightarrow \frac{i(a+\overline{a})i(b+\overline{b})}{(a-\overline{a})(b-\overline{b})}=-1$
$\Rightarrow ab+a\overline{b}+\overline{a}b+\overline{a}\overline{b}=ab-a\overline{b}-\overline{a}b+\overline{a}\overline{b}\therefore a\overline{b}+\overline{a}b=0$
Hence,option 'D' is correct.