Q86 / 200 · WBJEE 2013 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'D'.

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