Explanation

(a)
Let $\mathcal{z}_{1} = a + ib = (a,\ b)$
and $\mathcal{z}_{2} = c - id = (c,\ - d)$
where $a > 0$ and $d > 0$
Given, |$\mathcal{z}_{1}| = |\mathcal{z}_{2}|$
$\Rightarrow \ \ a^{2} + b^{2} = c^{2} + d^{2}$ …(i)
Now, $\frac{\mathcal{z}_{1} + \mathcal{z}_{2}}{\mathcal{z}_{1} - \mathcal{z}_{2}} = \frac{(a + ib) + (c - id)}{(a + ib) - (c - id)}$
$= \ \frac{(a + c) + i(b - d)}{(a - c) + i(b + d)}$
$= \frac{\lbrack(a + c) + i(b - d)\rbrack}{\lbrack(a - c) + i(b + d)\rbrack}\frac{\lbrack(a - c) - i(b + d)\rbrack}{\lbrack(a - c) - i(b + d)\rbrack}$
$= \ \frac{\left( a^{2} + b^{2} \right) - \left( c^{2} + d^{2} \right) - 2(ad + bc)i}{a^{2} + c^{2} - 2ac + b^{2} + d^{2} + 2bd}$
$= \ \frac{- (ad + bc)i}{a^{2} + b^{2} - ac + bd}\ \ \ \ \ \ \lbrack from\ Eq.(i)\rbrack$
$\therefore\ \ \frac{(\mathcal{z}_{1} + \mathcal{z}_{2})}{(\mathcal{z}_{1} - \mathcal{z}_{2})}$ is purely imaginary
Alternative , Assume any two complex number satisfying both conditions, $\mathcal{z}_{1} \neq \mathcal{z}_{2}$ and $\left| \mathcal{z}_{1} \right| = \left| \mathcal{z}_{2} \right|$
Let $\mathcal{z}_{1} = 2 + i,\ \mathcal{z}_{2} = 1 - 2i,$
$\therefore\ \ \frac{\mathcal{z}_{1} + \mathcal{z}_{2}}{\mathcal{z}_{1} - \mathcal{z}_{2}} = \frac{3 - i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i} = - \frac{10i}{10} = - i$
$\therefore$ It is purely imaginary