Answer
$ -\frac{3}{5}-\frac{4}{5}i$
Work Step by Step
Given
\begin{equation}
f(x)=\frac{x^2+11}{3-x}
\end{equation}
We want to find
\begin{equation}
f(4 i)
\end{equation}
Let $N(x) = x^2+11$ and $D(x) = 3-x$, then
\begin{equation}
\begin{aligned}
N\left(4i \right)&=(4i)^2+11\\
&=16i^2+11\\
&= 16(-1)+11\\
&= -16+11\\
&= -5\\
D\left(4i\right)&=3-4i\\
\end{aligned}
\end{equation}
It follows that:
\begin{equation}
\begin{aligned}
f(4 i)&= \frac{N(4i)}{D(4i)} \\
&=-\frac{5}{3-4i} \\
&= -\frac{5}{(3-4i)}\cdot \frac{(3+4i)}{(3+4i)}\\
& = -\frac{15+20i}{9+12i-12i-16i^2}\\
& = -\frac{15+20i}{9-16\cdot (-1)} \\
& = -\frac{15+20i}{9+16}\\
& = -\frac{15+20i}{25}\\
& =-\frac{15}{25}-\frac{20i}{25}\\
& = -\frac{3}{5}-\frac{4}{5}i
\end{aligned}
\end{equation}
We could have also used the fact that $(a-bi)(a+bi) = a^2+b^2$ to make the solution shorter.
\begin{equation}
\begin{aligned}
f(4 i)&= \frac{N(4i)}{D(4i)} \\
&=-\frac{5}{3-4i} \\
&= -\frac{5}{(3-4i)}\cdot \frac{(3+4i)}{(3+4i)}\\
& = -\frac{15+20i}{3^3+4^2}\\
& = -\frac{15+20i}{25} \\
& =-\frac{15}{25}-\frac{20i}{25}\\
& = -\frac{3}{5}-\frac{4}{5}i
\end{aligned}
\end{equation}
