Answer
$\frac{12}{25}+\frac{9}{25}i$
Work Step by Step
Given: $\displaystyle \frac{3}{4-3i}$
Multiply the numerator and denominator by the conjugate of the denominator, as follows:
$\frac{3}{4-3i}=\frac{3}{4-3i}\times\frac{4+3i}{4+3i}$
Since, $i^2=-1$ and $i=\sqrt{-1}$
Thus,
$=\frac{12+9i}{16-9(-1)}$
$=\frac{12}{25}+\frac{9}{25}i$
