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
- an ellipse
- a circle
- a parabola
- a straight line
The Correct Option is D
Solution and Explanation
Now, $\frac{z-1}{z-i}=\frac{(x+i y)-1}{(x+i y)-i}$
$=\frac{(x-1)+i y}{x+i(y-1)} \times \frac{x-i(y-1)}{x-i(y-1)}$
$=\frac{x(x-1)+i x y-i(x-1)(y-1)+y(y-1)}{x^{2}+(y-1)^{2}}$
$=\frac{\left(x^{2}+y^{2}-x-y\right)+i(x y-x y+y+x-1)}{\left(x^{2}+y^{2}+t^{-2 y}\right)}$
$=\left(\frac{x^{2}+y^{2}-x-y}{x^{2}+y^{2}-2 y+1}\right)+i\left(\frac{x+y-1}{x^{2}+y^{2}-2 y+1}\right) \ldots(1)$
Given, $\frac{z-1}{z-i}$ is real.
So, its imaginary part should be zero. i.e.,
$\frac{x+y-1}{x^{2}+y^{2}-2 y+1}=0$
$\Rightarrow x+y=1$
$\left(\because x^{2}+y^{2}-2 y+1 \neq 0\right)$
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Concepts Used:
Argand Plane
It is a set of 3 mutually perpendicular axes and a convenient way to represent a set of numbers (2 or 3) or a point in space.
Hence, we have a way to represent an imaginary number graphically. All we need to do is to find the real part and an imaginary part of it. Then, represent them on the two mutually perpendicular number lines. The point of intersection, as shown in the above diagram, is the origin of our Plane.
The formation of the Plane so formed is known as the Argand Plane and it is a convenient way to represent an imaginary number graphically.