Question

Let a and b be two non-zero complex numbers. If the lines $\overline{az}+\overline{az}+1=0$ and $\overline{bz}+\overline{bz}-1=0$ are mutually perpendicular, then a, b are connected by the relation

• 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 $a\overline{b}+\overline{a}b=0$

We know that multiplication of a complex number by i means rotation through an angle ${90}^{\circ },$ Replacing $z$ by $iz$ or $\overline{z}$ by $\overline{i}z=\overline{i}\overline{z}=-i\overline{z}$ in one line should give the other perpendicular line.
$\therefore a\left(\overline{i}z\right)+\overline{a}\left(iz\right)+1=0$
or $-ai\overline{z}+\overline{a}iz+1=0$
$ai\overline{z}-\overline{a}iz-1=0$ is same as
$b\overline{z}+\overline{b}z-1=0$
Comparing $\frac{ai}{b}=-\frac{\overline{a}i}{\overline{b}}$
or $a\overline{b}+\overline{a}b=0$