Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 5

The simple zero assumption gives the local factorizationwhere $k$ is holomorphic. In a small disc $k$ stays nonzero, so $q=g/k$ is holomorphic there. Expanding $q$ to first order,The residue is thereforeIf $g(a)=0$, the quotient has a removable singularity rather than a simple pole, and the same formula correctly gives zero. This is the residue formula for a quotient at a simple denominator zero.

The improper integral converges: its integrand is bounded near zero and at most $x^{-n}$ for $x\ge 1$, which is integrable because $n\ge 2$.

Set $\theta =\pi /n$, $\omega =e^{2i\theta }$ and $\zeta =e^{i\theta }$. Apply the residue theorem to $F(z)=1/(1+z^n)$ on the positively oriented sector contour consisting of the segment $0\to R$, the arc $R{e}^{it}$ for $0\le t\le 2\theta$, and the segment $R\omega \to 0$, with $R>1$. No pole lies on the boundary. Exactly one root of $z^n=-1$ is inside this sector, namely $\zeta$; it is a simple pole. Part (a) givesOn the returning ray, put $z=\omega x$ with $x$ decreasing from $R$ to zero. Since ${\omega }^n=1$, the two ray contributions areOn the arc, $|1+z^n|\ge R^n-1$, so the estimation lemma bounds its contour integral byThe residue theorem and the convergence already established now giveUsing $1-e^{2i\theta }=-2i{e}^{i\theta }\sin \theta$, we concludeThe sector also works for $n=2$, when it is the upper semicircle. Its angle is chosen so that $z^n$ has the same value on the two rays.

Factor the denominator as $(z^2+1)(z^2+4)$ and integrateover $[-R,R]$ followed by the upper semicircle, for $R>2$. The enclosed simple poles are $i$ and $2i$. The local quotient formula from part (a) givesOn this arc, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+1|\ge R^2-1$, $|z^2+4|\ge R^2-4$. Thus the estimation lemma givesOn the real line, $|F(x)|\le 1/[(x^2+1)(x^2+4)]$, so the improper integral is absolutely convergent. Hence the residue theorem yieldsTaking real parts is justified by absolute convergence. ThereforeThe imaginary part vanishes as a further check, since the sine integrand is odd.