Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 4

The Cauchy derivative formula says that if $f$ is holomorphic on a neighbourhood of a closed disk and $z_0$ is inside its positively oriented boundary circle $\gamma$, then for $n\ge 0$,For $n=0$ this is the Cauchy integral formula. The same formula holds for a positively oriented simple closed contour when $f$ is holomorphic on a neighbourhood of the contour and its enclosed region.

In the present problem the unit circle and its closed interior lie inside $|z|<R$. Parametrize it by $z=e^{it}$, $0\le t\le 2\pi$, so $dt=dz/(iz)$. For a positive integer $n$,The first contour integral is zero by the Cauchy integral theorem, since $f(z)z^{n-1}$ is holomorphic on the disk; the second is evaluated by the Cauchy derivative formula. ThereforeTaking $n=1$ gives the requested $\pi f^{\prime }(0)$.

The expression is $F(z)=e^{z^2+1/z}=e^{z^2}{e}^{1/z}$, which is holomorphic for every $z\ne 0$. At zero,has infinitely many nonzero negative-power terms, so it has an essential singularity. Multiplication by the nonvanishing entire function $e^{z^2}$ preserves that classification: if $F$ had a removable singularity or a pole, multiplying by $e^{-z^2}$ would give the same type for $e^{1/z}$, a contradiction. Hence

The only possible finite singularities are $z_{\pm }=\pm i\pi /2$, the zeros of $4z^2+{\pi }^2$. They are simple zeros of the denominator, because its derivative $8z_{\pm }$ is nonzero. They are also simple zeros of the numerator: $\cosh z_{\pm }=0$ and $\sinh z_{\pm }=\pm i\ne 0$.

The common simple factors cancel, so both points are removable singularities, with extension valuesThere are no other finite singularities; after inserting these two values the function is entire.

For $z\ne 0$, the denominator vanishes exactly when $1/z=(2k+1)\pi i$. Thus setEach denominator zero is simple, sinceEvery $z_k$ is purely imaginary and nonzero. The identity $\sin (iy)=i\sinh y$ for real $y$ shows $\sin z_k\ne 0$, so there is no cancellation. Each $z_k$ is consequently a simple pole, with residue $z_k^2\sin z_k$.

These poles accumulate at zero. ThereforeIn particular, zero has no punctured neighbourhood on which the function is holomorphic, so it cannot be classified as a removable singularity, a pole, or an isolated essential singularity. There are no other finite singularities.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence