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

Given: $\frac{1}{4+3 i}$

Now, rationalizing

$=\frac{1}{4+3 i} \times \frac{4-3 i}{4-3 i}$

$=\frac{4-3 i}{(4+3 i)(4-3 i)} \ldots(\mathrm{i})$

Now, we know that,

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

So, eq. (i) become

$=\frac{4-3 i}{(4)^{2}-(3 i)^{2}}$

$=\frac{4-3 i}{16-9 i^{2}}$

$=\frac{4-3 i}{16-9(-1)}\left[\because i^{2}=-1\right]$

$=\frac{4-3 i}{16+9}$

$=\frac{4-3 i}{25}$

$=\frac{4}{25}-\frac{3}{25} i$