Answer
$\frac{12}{25}+\frac{9}{25}i$
Work Step by Step
$\displaystyle \frac{3}{4-3i}$
We multiply the numerator and denominator by the conjugate of the denominator:
$=\frac{3}{4-3i}\cdot\frac{4+3i}{4+3i}$
We use the fact that $i^2=-1$ (because $i=\sqrt{-1}$)
$=\frac{12+9i}{16-9*-1}$
$=\frac{12}{25}+\frac{9}{25}i$
