Question

If $\overline{z}={\overline{z}}_0+A(z-z_0)$, where A is a constant, then prove that the locus of z is a straight line.

Answer 1

$\overline{z}={\overline{z}}_0+A(z-z_0)$
$Az-\overline{z}-A{z}_0+{\overline{z}}_0=0$ (1)
$\overline{A}\overline{z}-z-\overline{A}{\overline{z}}_0+z_0=0$ (2)
Adding (1) and (2), we get
$(A-1)z+(\overline{A}-1)\overline{z}-(A{z}_0+\overline{A}{\overline{z}}_0)+z_0+{\overline{z}}_0=0$
This is of the form $\overline{a}z+a\overline{z}+b=0$, where $a=\overline{A}-1$ and $b=-(A{z}_0+\overline{A}{\overline{z}}_0)+z_0+{\overline{z}}_0\in R.$
Hence, locus of z is a straight line.
Ans: 1