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

The correct option is D A straight line

Given $z=x+iy$

$i=\sqrt{-1}$

$\Rightarrow \frac{z-1}{z-i}=k$ ( real )

$\Rightarrow \frac{(x-1)+iy}{x+(y-1)i}=\frac{(x-1)+iy}{x+(y-1)i}\times \frac{x-(y-1)i}{x-(y-1)i}$

$=\frac{x^2-x+xyi-(xy-x-y+1)i+(y^2-y)}{x^2-{(y-1)}^2}$

For real , imaginary part $=0$

$\Rightarrow \frac{xy-xy+x+y-1}{x^2-{(y-1)}^2}=0$

$\Rightarrow x+y=1$

$\therefore$ Straight line

$(x,y)$ lies on straight line.

Hence, the answer is a straight line.