Question

Consider the equation $az+b\overline{z}+c=0$, where $a,b,cϵZ$. If $|a|-|b|\ne 0$ and $\overline{a}c=b\overline{c}$, then $az+b\overline{z}+c=0$ represents

• A. An ellipse
• B. A circle
• C. A point
• D. A straight line

Answer 1

The correct option is D A straight line

$a\overline{z}+b\overline{z}+c=0$ (1)
or $\overline{a}z+\overline{b}z+\overline{c}=0$ (2)
Eliminating $\overline{z}$ from (1) and (2), we geet
$z=\frac{c\overline{a}-b\overline{c}}{|b{|}^2-|a{|}^2}$
If $|a|\ne |b|$, then $z$ represents one point on the Argand plane. If $|a|=|b|$ and $\overline{a}c\ne b\overline{c}$, then no such $z$ exists. Adding (1) and (2),
$(\overline{a}+b)\overline{z}+(a+\overline{b})z+(c+\overline{c})=0$
This is of the form $A\overline{z}+\overline{A}z+B=0$, where $B=c+\overline{c}=0$ is real.
Hence, locus of $z$ is a straight line.