Express each of the following in the form (a + ib)

Solution:

Given: $\frac{-2+5 i}{3-5 i}$

Now, rationalizing

$=\frac{-2+5 i}{3-5 i} \times \frac{3+5 i}{3+5 i}$

$=\frac{(-2+5 i)(3+5 i)}{(3-5 i)(3+5 i)} \ldots(\mathrm{i})$

Now, we know that,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

So, eq. (i) become

$=\frac{(-2+5 i)(3+5 i)}{(3)^{2}-(5 i)^{2}}$

$=\frac{-2(3)+(-2)(5 i)+5 i(3)+5 i(5 i)}{9-25 i^{2}}$

$=\frac{-6-10 i+15 i+25 i^{2}}{9-25(-1)}\left[\because i^{2}=-1\right]$

$=\frac{-6+5 i+25(-1)}{9+25}$

$=\frac{-31+5 i}{34}$

$=-\frac{31}{34}+\frac{5}{34} i$