Answer
$\dfrac{2-3i}{1-2i}=\dfrac{8}{5}+\dfrac{1}{5}i$
Work Step by Step
$\dfrac{2-3i}{1-2i}$
Multiply the fraction by $\dfrac{1+2i}{1+2i}$:
$\Big(\dfrac{2-3i}{1-2i}\Big)\Big(\dfrac{1+2i}{1+2i}\Big)=\dfrac{2+4i-3i-6i^{2}}{1-(2i)^{2}}=\dfrac{2+i-6i^{2}}{1-4i^{2}}=...$
Substitute $i^{2}$ with $-1$:
$...=\dfrac{2+i-6(-1)}{1-4(-1)}=\dfrac{8+i}{5}=\dfrac{8}{5}+\dfrac{1}{5}i$
