P is a point denoting z in the Argand diagram | vedclass

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(C) Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.To make this purely imaginary,we multiply by the conjugate of the denomin

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the points on the circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$,excluding the points $(1, 0)$ and $(0, 1)$

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(C) Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.To make this purely imaginary,we multiply by the conjugate of the denominator: $\frac{(x + i(y-1))((x-1) - iy)}{(x-1)^2 + y^2}$.The real part is $\frac{x(x-1) + y(y-1)}{(x-1)^2 + y^2} = 0$.This implies $x^2 - x + y^2 - y = 0$,which is $x^2 + y^2 - x - y = 0$.Completing the square,we get $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2$.This represents a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$.Since the expression is undefined at $z = 1$ (i.e.,$(1, 0)$) and the numerator is zero at $z = i$ (i.e.,$(0, 1)$),these points must be excluded.

$P$ is a point denoting $z$ in the Argand diagram. If $\frac{z-i}{z-1}$ is always purely imaginary,then the locus of $P$ is

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