If z is a complex number such that frac{z-1}{ | vedclass

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(D) Let $z = x + iy$. Then $\frac{z-1}{z-i} = \frac{(x-1) + iy}{x + i(y-1)}$.Multiplying numerator and denominator by the conjugate of the denominator

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$4 r^2$

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(D) Let $z = x + iy$. Then $\frac{z-1}{z-i} = \frac{(x-1) + iy}{x + i(y-1)}$.Multiplying numerator and denominator by the conjugate of the denominator: $\frac{((x-1) + iy)(x - i(y-1))}{x^2 + (y-1)^2}$.For the expression to be purely imaginary,the real part must be zero: $x(x-1) + y(y-1) = 0$.This simplifies to $x^2 - x + y^2 - y = 0$,or $(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}$.Comparing this with the equation of a circle $(x - \alpha)^2 + (y - \beta)^2 = r^2$,we get $\alpha = \frac{1}{2}$,$\beta = \frac{1}{2}$,and $r^2 = \frac{1}{2}$.Thus,$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{1/2}{1/2} + \frac{1/2}{1/2} = 1 + 1 = 2$.Since $r^2 = \frac{1}{2}$,we have $2 = 4r^2$. Therefore,the correct option is $4r^2$.

If $z$ is a complex number such that $\frac{z-1}{z-i}$ is purely imaginary and the locus of $z$ represents a circle with centre $(\alpha, \beta)$ and radius $r$,then $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=$

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