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

Given: $\frac{3+4 i}{4+5 i}$

Now, rationalizing

$=\frac{3+4 i}{4+5 i} \times \frac{4-5 i}{4-5 i}$

$=\frac{(3+4 i)(4-5 i)}{(4+5 i)(4-5 i)}$

Now, we know that

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

So, eq. (i) become

$=\frac{(3+4 i)(4-5 i)}{(4)^{2}-(5 i)^{2}}$

$=\frac{3(4)+3(-5 i)+4 i(4)+4 i(-5 i)}{16-25 i^{2}}$

$=\frac{12-15 i+16 i-20 i^{2}}{16-25(-1) \quad\left[\because i^{2}=-1\right]}$

$=\frac{12+i-20(-1)}{16+25}$

$=\frac{12+i+20}{41}$

$=\frac{32+i}{41}$

$=\frac{32}{41}+\frac{1}{41} i$