Explanation

(a)
Let $\mathcal{z =}x + iy$
$\therefore\ \frac{\mathcal{z +}i}{\mathcal{z} + 2} = \frac{x + iy + i}{x + iy + 2} = \frac{x + i(y + 1)}{(x + 2) + iy}$
$= \frac{\left\lbrack x + i(y + 1) \right\rbrack \times \lbrack(x + 2) - iy\rbrack}{\left\lbrack (x + 2) + iy \right\rbrack \times \lbrack(x + 2) - iy\rbrack}$
$= \left\lbrack \frac{x^{2} + 2x + y^{2} + y}{(x + 2)^{2} + y^{2}} \right\rbrack + i\left\lbrack \frac{(y + 1)(x + 2) - xy}{(x + 2)^{2} + y^{2}} \right\rbrack$
Since, it is purely imaginary, therefore real part must be equal to zero
$\therefore\ \frac{x^{2} + y^{2} + 2x + y}{(x + 2)^{2} + y^{2}} = 0$
$\Rightarrow \ x^{2} + y^{2} + 2x + y = 0$
It represents the equation of circle and its radius
$= \sqrt{1 + \frac{1}{4} - 0} = \frac{\sqrt{5}}{2}$
Therefore, locus of $\mathcal{z}$ in argand diagram is a circle of radius $\frac{\sqrt{5}}{2}$