Answer
The simplified form of the expression $\frac{{{x}^{2}}+19}{2-x}$ for $x=3i$ is $\frac{20}{13}+\frac{30}{13}i$.
Work Step by Step
Consider the expression, $\frac{{{x}^{2}}+19}{2-x}$
Substitute $x=3i$ in the expression $\frac{{{x}^{2}}+19}{2-x}$.
$\begin{align}
& \frac{{{x}^{2}}+19}{2-x}=\frac{{{\left( 3i \right)}^{2}}+19}{2-\left( 3i \right)} \\
& =\frac{9{{i}^{2}}+19}{2-3i}
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \frac{{{x}^{2}}+19}{2-x}=\frac{9\left( -1 \right)+19}{2-3i} \\
& =\frac{-9+19}{2-3i} \\
& =\frac{10}{2-3i}
\end{align}$
Multiply by complex conjugate of the denominator in the numerator and the denominator.
\[\frac{{{x}^{2}}+19}{2-x}=\frac{10}{\left( 2-3i \right)}\cdot \frac{\left( 2+3i \right)}{\left( 2+3i \right)}\]
Distribute $10$ within the parentheses in numerator and use the FOIL method in the denominator.
\[\begin{align}
& \frac{{{x}^{2}}+19}{2-x}=\frac{10\left( 2+3i \right)}{\left( 2-3i \right)\left( 2+3i \right)} \\
& =\frac{20+30i}{4+6i-6i-9{{i}^{2}}} \\
& =\frac{20+30i}{4-9{{i}^{2}}}
\end{align}\]
Replace the value ${{i}^{2}}=-1$.
\[\begin{align}
& \frac{{{x}^{2}}+19x}{2-x}=\frac{20+30i}{4-9\left( -1 \right)} \\
& =\frac{20+30i}{4+9} \\
& =\frac{20+30i}{13} \\
& =\frac{20}{13}+\frac{30}{13}i
\end{align}\]
Therefore, the simplified form of the expression $\frac{{{x}^{2}}+19}{2-x}$ for $x=3i$ is $\frac{20}{13}+\frac{30}{13}i$.
