Practise exact division and combine complex fractions

Multiply by the denominator’s conjugate:

$\begin{aligned}&\frac{2+3i}{3-4i}\\ & =\frac{(2+3i)(3+4i)}{3^2+4^2}\\ & =\frac{6+8i+9i+12i^2}{25}\\ & =\frac{-6+17i}{25}.\end{aligned}$

Use $4+i$ to make the denominator real:

$\begin{aligned}&\frac{3-i}{4-i}\\ & =\frac{(3-i)(4+i)}{4^2+1^2}\\ & =\frac{12+3i-4i-i^2}{17}\\ & =\frac{13-i}{17}.\end{aligned}$

Use $1+3i$ as the conjugate factor:

$\begin{aligned}&\frac{-2+5i}{1-3i}\\ & =\frac{(-2+5i)(1+3i)}{1^2+3^2}\\ & =\frac{-2-6i+5i+15i^2}{10}\\ & =\frac{-17-i}{10}.\end{aligned}$

Realise each denominator. Their imaginary parts cancel:

$\begin{aligned}\frac1{1+3i}+\frac1{1-3i}&=\frac{1-3i}{10}+\frac{1+3i}{10}\\&=\frac2{10}=\frac15.\end{aligned}$

First simplify the quotient, then add corresponding components:

$\begin{aligned}-2+i+\frac{3i}{2-i}&=-2+i+\frac{3i(2+i)}5\\&=-2+i+\frac{-3+6i}5\\&=-\frac{13}5+\frac{11}5i.\end{aligned}$

The two real denominators are both 25:

$\begin{aligned}\frac{1-2i}{3+4i}+\frac{-2-i}{4+3i}&=\frac{(1-2i)(3-4i)}{25}+\frac{(-2-i)(4-3i)}{25}\\&=\frac{3-4i-6i+8i^2}{25}+\frac{-8+6i-4i+3i^2}{25}\\&=\frac{-5-10i}{25}+\frac{-11+2i}{25}\\&=-\frac{16}{25}-\frac8{25}i.\end{aligned}$