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)

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