Answer
$-\dfrac{24}{25} + \dfrac{32}{25}i$
Work Step by Step
Rationalize the denominator by multiplying the conjugate of the denominator, which is $4+3i$, to both the numerator and the denominator to obtain:
$=\dfrac{8i(4+3i)}{(4-3i)(4+3i)}
\\=\dfrac{32i+24i^2}{(4-3i)(4+3i)}$
Use the rule $(a-b)(a+b) = a^2-b^2$ to obtain:
$=\dfrac{32i+24i^2}{4^2-(3i)^2}
\\=\dfrac{32i+24i^2}{16-9i^2}$
Use the fact that $i^2=-1$ to obtain:
$=\dfrac{32i+24(-1)}{16-9(-1)}
\\=\dfrac{32i-24}{16+9}
\\=\dfrac{-24+32i}{25}
\\=-\dfrac{24}{25} + \dfrac{32}{25}i$
