Answer
$\dfrac{20}{13}+\dfrac{30}{13}i$
Work Step by Step
$\dfrac{x^{2}+19}{2-x}$ for $x=3i$
Substitute $x$ by $3i$:
$\dfrac{(3i)^{2}+19}{2-3i}=\dfrac{9i^{2}+19}{2-3i}=...$
Substitute $i^{2}$ by $-1$:
$...=\dfrac{9(-1)+19}{2-3i}=\dfrac{-9+19}{2-3i}=\dfrac{10}{2-3i}=...$
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator:
$...=\dfrac{10}{2-3i}\cdot\dfrac{2+3i}{2+3i}=\dfrac{10(2+3i)}{2^{2}-(3i)^{2}}=\dfrac{20+30i}{4-9i^{2}}=...$
Substitute $i^{2}$ by $-1$ again and simplify:
$...=\dfrac{20+30i}{4-9(-1)}=\dfrac{20+30i}{4+9}=\dfrac{20+30i}{13}=\dfrac{20}{13}+\dfrac{30}{13}i$
