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WBJEE · Maths · Complex Number

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

  1. A an ellipse
  2. B a circle
  3. C a parabola
  4. D a straight line
Verified Solution

Answer & Solution

Correct Answer

(D) a straight line

Step-by-step Solution

Detailed explanation

Given, \(z=x+i y\) Now, \(\frac{z-1}{z-i}=\frac{(x+y)-1}{(x+i y)-i}\) \(=\frac{(x-1)+i y}{x+i(y-1)} \times \frac{x-i(y-1)}{x-i(y-1)}\) \(=\frac{x(x-1)+i x y-i(x-1)(y-1)+y(y-1)}{x^{2}+(y-1)^{2}}\)…