= Solution
\b[Singularities.] The <complex exponential> factors as $F(z)=e^{z/2}e^{1/(2z)}$. The second factor has an <essential singularity> at zero, and multiplication by the nonvanishing <holomorphic function> $e^{z/2}$ does not remove it. There are no other finite <singularities>. On the <Riemann sphere>, infinity is also an <essential singularity>, since $F(1/\zeta)=F(\zeta)$.
\b[Symmetry and integral formula.] The identity $F(1/z)=F(z)$ and uniqueness of the <Laurent series> on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The <Laurent coefficient> formula, on the positively oriented <unit circle>, gives
$$
F_k=\frac{1}{2\pi i}\oint_{|z|=1}F(z)z^{-k-1}\,dz
=\frac{1}{2\pi}\int_0^{2\pi}e^{\cos\theta}e^{-ik\theta}\,d\theta.
$$
The imaginary part vanishes by the substitution $\theta\mapsto2\pi-\theta$. Hence
$$
\boxed{F_k=F_{-k}=\frac{1}{2\pi}\int_0^{2\pi}e^{\cos\theta}\cos(k\theta)\,d\theta.}
$$
\b[Polynomial contour integral.] For an integer $n\geq0$ and integer $p$, the <binomial theorem> gives $(1+z^2)^n=\sum_{j=0}^n\binom nj z^{2j}$. The <residue theorem> selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement is
$$
\boxed{\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z^2)^n}{z^{n+p+1}}\,dz=
\begin{cases}
\displaystyle\frac{n!}{((n-p)/2)!((n+p)/2)!},&n+p\text{ even and }|p|\leq n,\\
0,&\text{otherwise}.
\end{cases}}
$$
The printed nonnegative-integer condition alone misses $j\leq n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.
\b[Explicit coefficients.] Both <Taylor series> in
$$
e^{z/2}e^{1/(2z)}=\sum_{r,s\geq0}\frac{z^{r-s}}{2^{r+s}r!s!}
$$
converge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\geq0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:
$$
\boxed{F_k=\sum_{m=0}^\infty\frac{1}{2^{|k|+2m}m!(|k|+m)!}=I_{|k|}(1).}
$$
Here $I_j$ is a <modified Bessel function>. The printed series is directly meaningful for $k\geq0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.
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