## Mar 2, 2018

$\frac{3}{4}$.

#### Explanation:

Let, z=x+iy; x,y in RR.

$\therefore \frac{i z - 2}{z - i} = \frac{i \left(x + i y\right) - 2}{x + i y - i} = \frac{- y - 2 + i x}{x + i \left(y - 1\right)}$,

$= \frac{- y - 2 + i x}{x + i \left(y - 1\right)} \times \frac{x - i \left(y - 1\right)}{x - i \left(y - 1\right)}$,

$= \frac{x \left(- y - 2\right) + i {x}^{2} + i \left(y - 1\right) \left(y + 2\right) - {i}^{2} x \left(y - 1\right)}{{x}^{2} - {i}^{2} {\left(y - 1\right)}^{2}}$,

$= \frac{x \left(- y - 2 + y - 1\right) + i \left\{{x}^{2} + \left(y - 1\right) \left(y + 2\right)\right\}}{{x}^{2} + {\left(y - 1\right)}^{2}}$.

Clearly, $I m \left(\frac{i z - 2}{z - i}\right) = \frac{{x}^{2} + \left(y - 1\right) \left(y + 2\right)}{{x}^{2} + {\left(y - 1\right)}^{2}}$,

$= - 1. \ldots \ldots \ldots \ldots \text{[because, Given]}$,

$\Rightarrow \left\{{x}^{2} + \left(y - 1\right) \left(y + 2\right)\right\} + \left\{{x}^{2} + {\left(y - 1\right)}^{2}\right\} = 0$,

$\Rightarrow 2 {x}^{2} + {y}^{2} + y - 2 + {y}^{2} - 2 y + 1 = 0 , \mathmr{and}$,

$2 {x}^{2} + 2 {y}^{2} - y = 1$.

$\therefore {x}^{2} + {y}^{2} - \frac{1}{2} y = \frac{1}{2} , \mathmr{and}$,

${x}^{2} + {\left(y - \frac{1}{4}\right)}^{2} = {\left(\frac{1}{4}\right)}^{2} + \frac{1}{2} = \frac{9}{16} = {\left(\frac{3}{4}\right)}^{2}$.

This represents a circle with centre at $\left(0 , \frac{1}{4}\right)$ and radius

$\frac{3}{4}$.