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
- A an ellipse
- B a circle
- C a parabola
- D a straight line
Answer & Solution
Correct Answer
(D) a straight line
Step-by-step Solution
Detailed explanation
Given, \(z=x+i y\) Now, \(\frac{z-1}{z-i}=\frac{(x+y)-1}{(x+i y)-i}\) \(=\frac{(x-1)+i y}{x+i(y-1)} \times \frac{x-i(y-1)}{x-i(y-1)}\) \(=\frac{x(x-1)+i x y-i(x-1)(y-1)+y(y-1)}{x^{2}+(y-1)^{2}}\)…
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