#### Answer

$-\dfrac{3}{5}-\dfrac{4}{5}i$

#### Work Step by Step

$\dfrac{x^{2}+11}{3-x}$ for $x=4i$
Substitute $x$ by $4i$:
$\dfrac{(4i)^{2}+11}{3-4i}=\dfrac{16i^{2}+11}{3-4i}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{16(-1)+11}{3-4i}=\dfrac{-16+11}{3-4i}=\dfrac{-5}{3-4i}=...$
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator:
$...=\dfrac{-5}{3-4i}\cdot\dfrac{3+4i}{3+4i}=\dfrac{-5(3+4i)}{3^{2}-(4i)^{2}}=\dfrac{-15-20i}{9-16i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{-15-20i}{9-16(-1)}=\dfrac{-15-20i}{9+16}=\dfrac{-15-20i}{25}=-\dfrac{15}{25}-\dfrac{20}{25}i=...$
$...=-\dfrac{3}{5}-\dfrac{4}{5}i$