Simplify each of the following and express it in the form (a + ib) :

Solution:

Given: $(4-3 i)^{-1}$

We can re- write the above equation as

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

Now, rationalizing

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

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

Now, we know that,

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

So, eq. (i) become

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

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

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

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

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

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