Question

Let $z_1$ and $z_2$ are fixed complex numbers, the locus of $z$ so that
$\frac{z-z_1}{z-z_2}=e^{i\theta },\theta \in R$
is

• A. A circle
• B. A semicircle
• C. A straight line
• D. Perpendicular Bisector of segment joining $z_1$ and $z_2$

Answer 1

Answer: (C), (D)

Perpendicular Bisector of segment joining $z_1$ and $z_2$
A straight line

Given:$\frac{z-z_1}{z-z_2}=e^{i\theta }\theta \in R$

Taking modulus

$\Rightarrow \left|\begin{array}{c}z-z_1\end{array}\right|=\left|\begin{array}{c}{e}^{i\theta }\end{array}\right|\left|\begin{array}{c}z-z_2\end{array}\right|$

As $\left|\begin{array}{c}{e}^{i\theta }\end{array}\right|=1$

$\Rightarrow \left|\begin{array}{c}z-z_1\end{array}\right|=\left|\begin{array}{c}z-z_2\end{array}\right|$

Hence the locus of a point equidistant from two given points is perpendicular bisector of the line joining those points