If z=x+iy, where x and y are real numbers and | vedclass

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(D) Given $z=x+iy$. Then $\frac{z-1}{z-i} = \frac{(x-1)+iy}{x+i(y-1)}$. To make this expression real, we multiply the numerator and denominator by the

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a straight line

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(D) Given $z=x+iy$. Then $\frac{z-1}{z-i} = \frac{(x-1)+iy}{x+i(y-1)}$. To make this expression real, we multiply the numerator and denominator by the conjugate of the denominator: $\frac{(x-1)+iy}{x+i(y-1)} 	imes \frac{x-i(y-1)}{x-i(y-1)} = \frac{x(x-1) - i(x-1)(y-1) + ixy + y(y-1)}{x^2 + (y-1)^2}$. The imaginary part of this expression is $\frac{xy - (x-1)(y-1)}{x^2 + (y-1)^2} = \frac{xy - (xy - x - y + 1)}{x^2 + (y-1)^2} = \frac{x+y-1}{x^2 + (y-1)^2}$. For the expression to be real, the imaginary part must be zero: $\frac{x+y-1}{x^2 + (y-1)^2} = 0 \implies x+y-1 = 0$ (where $z 
eq i$). This is the equation of a straight line.

If $z=x+iy$, 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

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