Past exam of the short options paper of the physics Prelims of the University of Oxford / 2019 / S1 / 2

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

If $f(z)=\sum a_j{z}^j$, then $f(-z)=\sum (-1{)}^j{a}_j{z}^j$ and its residue at zero is $-a_{-1}$. The circle $|z-1|=2$ encloses zero and no other finite singularity. Thus the residue theorem givesfor positive orientation.