Answer
$-\dfrac{12}{13} + \dfrac{18}{13}i$
Work Step by Step
Multiply both the numerator and the denominator by the conjugate of the denominator, which is $3+2i$, to obtain:
$=\dfrac{6i(3+2i)}{(3-2i)(3+2i)}
\\=\dfrac{18i+12i^2}{(3-2i)(3+2i)}$
Simplify using the rule $(a-b)(a+b)=a^2-b^2$ to obtain:
$=\dfrac{18i+12i^2}{3^2-(2i)^2}
\\=\dfrac{18i+12i^2}{9-4i^2}$
Use the fact that $i^2=-1$ to obtain:
$=\dfrac{18i+12(-1)}{9-4(-1)}
\\=\dfrac{18i-12}{9+4}
\\=\dfrac{-12+18i}{13}
\\=-\dfrac{12}{13} + \dfrac{18}{13}i$
