Answer
$\dfrac{7}{5}+\dfrac{31}{5}i$
Work Step by Step
We want to eliminate imaginary numbers in the denominators of a rational expression.
To do this, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.
The conjugate of the denominator is $2 + i$ so rationalize the denominator by multiplying $2+i$ to both the numerator and the denominator:
$=\dfrac{11i + 9}{2 - i} \cdot \dfrac{2 + i}{2 + i}$
$=\dfrac{(11i + 9)(2 + i)}{(2 - i)(2 + i)}$
Use the FOIL method to distribute terms, and use the fact that $(a-b)(a+b)=a^2-b^2$ and $i^2=-1$ to obtain:
$=\dfrac{22i + 11i^2 + 18 + 9i}{2^2- i^2}$
$=\dfrac{31i + 11i^2 + 18}{4 - i^2}$
$=\dfrac{31i + 11(-1) + 18}{4 - (-1)}$
$=\dfrac{31i - 11 + 18}{4 + 1}$
$=\dfrac{31i + 7}{5}$
