Answer
$\dfrac{9}{5} + \dfrac{18}{5}i$
Work Step by Step
Multiply both the numerator and the denominator by the conjugate of the denominator, which is $1+2i$, to obtain:
$=\dfrac{9(1+2i)}{(1-2i)(1+2i)}
\\=\dfrac{9+18i}{(1-2i)(1+2i)}$
Simplify using the rule $(a-b)(a+b)=a^2-b^2$ to obtain:
$=\dfrac{9+18i}{1^2-(2i)^2}
\\=\dfrac{9+18i}{1-4i^2}$
Use the fact that $i^2=-1$ to obtain:
$=\dfrac{9+18i}{1-4(-1)}
\\=\dfrac{9+18i}{1+4}
\\=\dfrac{9+18i}{5}
\\=\dfrac{9}{5} + \dfrac{18}{5}i$
