Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 13A (Complex Analysis or Complex Methods)

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

Using the Taylor series of the exponential and series division,Thus the first three nonzero Laurent terms areThe singularities occur at $z=k\pi i$. The nearest nonzero ones are $\pm \pi i$, so this Laurent series is valid on $0<|z|<\pi$.

The function $f$ has a simple pole of residue $1/2$ at every $z=k\pi i$, because $(e^{2z}-1{)}^{\prime }=2$ there. At $z=0$, the term $1/(2z)$ in $g$ cancels the principal part of $f$. For $1\le k\le m$, the summandhas residue $1/2$ at each of $z=\pm k\pi i$, so it cancels the corresponding pole of $f$. Therefore all possible singularities of $F$ in $|z|<(m+1)\pi$ are removable singularities: they are $0$ and $\pm k\pi i$ for $1\le k\le m$. After filling them in, $F$ is analytic throughout that disc.

Write the function from the preceding part as $F_m$. Its expansion at zero isHence the residue theorem givesChoose $R=R_m=(m+\frac{1}{2})\pi$. This circle stays a distance at least $\pi /2$ from every pole of $f$, so $f$ is bounded uniformly on it. MoreoverThus ${\max }_{C_{R_m}}|F_m|=O(\log m)$, and the ML inequality yieldsLetting $m\to \infty$ solves the Basel problem: