Question

If $\overline{z}=\overline{z_0}+A(z-z_0)$, where $A$ and $z_0$ are fixed complex numbers, then the locus of $z$ is a

• A. parabola
• B. hyperbola
• C. circle
• D. straight line

Answer 1

The correct option is D straight line

$\overline{z}=\overline{z_0}+A(z-z_0)$
$Az-\overline{z}-A{z}_0+\overline{z_0}=0\dots \left(1\right)$
$\overline{A}\overline{z}-z-\overline{A{z}_0}+z_0=0\dots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$,
$(A-1)z+(\overline{A}-1)\overline{z}-(A{z}_0+\overline{A{z}_0})+z_0+\overline{z_0}=0$
This is of the form $\overline{\alpha }z+\alpha \overline{z}+k=0$, where $\alpha =\overline{A}-1$ and $k=-(A{z}_0+\overline{A{z}_0})+z_0+\overline{z_0}\in R.$

Hence, locus of $z$ is a straight line.