(a)Let ${z_1} = a + ib = (a,b)$and ${z_2} = c – id = (c, – d)$
Where $a > 0$and $d > 0$ ……$(i)$
Then $|{z_1}| = |{z_2}|$ $⇒$  ${a^2} + {b^2} = {c^2} + {d^2}$
Now $\frac{{{z_1} + {z_2}}}{{{z_1} – {z_2}}} = \frac{{(a + ib) + (c – id)}}{{(a + ib) – (c – id)}}$
$ = \frac{{[(a + c) + i(b – d)][(a – c) – i(b + d)]}}{{[(a – c) + i(b + d)][(a – c) – i(b + d)]}}$
$ = \frac{{({a^2} + {b^2}) – ({c^2} + {d^2}) – 2(ad + bc)i}}{{{a^2} + {c^2} – 2ac + {b^2} + {d^2} + 2bd}}$ $\frac{{ – (ad + bc)i}}{{{a^2} + {b^2} – ac + bd}}$[using $(i)$]

$\therefore $$\frac{{({z_1} + {z_2})}}{{({z_1} – {z_2})}}$ is purely imaginary.

However if $ad + bc = 0$, then $\frac{{({z_1} + {z_2})}}{{({z_1} – {z_2})}}$ will be equal to zero. According to the conditions of the equation, we can have $ad + bc = 0$
Trick : Assume any two complex numbers satisfying both conditions i.e., ${z_1} \ne {z_2}$and $|{z_1}|\, = \,|{z_2}|$
Let ${z_1} = 2 + i,{z_2} = 1 – 2i,$

$\therefore \,\,\frac{{{z_1} + {z_2}}}{{{z_1} – {z_2}}} = \frac{{3 – i}}{{1 + 3i}} = – i$
Hence the result.