Question

Suppose $z = x + iy$ where $x$ and $y$ are real numbers and $i = \sqrt{-1}$. The points $(x, y)$ for which $\frac{z - 1}{z - i}$ is real, lie on

• A. an ellipse
• B. a circle
• C. a parabola
• D. a straight line

Answer 1

Answer: (D)

a straight line

Answer 2

Given, $z=x+i y$
Now, $\frac{z-1}{z-i}=\frac{(x+i y)-1}{(x+i y)-i}$
$=\frac{(x-1)+i y}{x+i(y-1)} \times \frac{x-i(y-1)}{x-i(y-1)}$
$=\frac{x(x-1)+i x y-i(x-1)(y-1)+y(y-1)}{x^{2}+(y-1)^{2}}$
$=\frac{\left(x^{2}+y^{2}-x-y\right)+i(x y-x y+y+x-1)}{\left(x^{2}+y^{2}+t^{-2 y}\right)}$
$=\left(\frac{x^{2}+y^{2}-x-y}{x^{2}+y^{2}-2 y+1}\right)+i\left(\frac{x+y-1}{x^{2}+y^{2}-2 y+1}\right) \ldots(1)$
Given, $\frac{z-1}{z-i}$ is real.
So, its imaginary part should be zero. i.e.,
$\frac{x+y-1}{x^{2}+y^{2}-2 y+1}=0$
$\Rightarrow x+y=1$
$\left(\because x^{2}+y^{2}-2 y+1 \neq 0\right)$