Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 3

For a piecewise continuously differentiable contour $\gamma :[a,b]\to U$, define its contour integral byadding the integrals over the finitely many smooth pieces if necessary. Continuity of $f$ along the contour makes each integrand integrable. The derivative supplies both the direction and the complex line element; replacing it by its modulus would give a different integral with respect to arc length.

For $\varphi (t)=r{e}^{it}$, ${\varphi }^{\prime }(t)=ir{e}^{it}$ and $|{\varphi }^{\prime }(t)|=r$. The triangle inequality for an integral applied to the parametrized contour integral givesThis is the circular-contour instance of the estimation lemma.

Use the continuous argument $0<\arg z<2\pi$ on the positive-axis slit plane and defineThis analytic logarithm on the positive-axis slit plane is holomorphic, satisfies $e^{L(z)}=z$ and $L^{\prime }(z)=1/z$, and has $L(-1)=i\pi$. That value fixes the branch uniquely on the connected domain. For positive real $x$, the two boundary limits areThe jump, rather than a pole of the integrand, is what the keyhole argument exploits.

The Cauchy integral theorem for a star-shaped domain says: if $U$ is open and star-shaped and $F$ is holomorphic on $U$, then ${\int }_{\gamma }F(z)dz=0$ for every closed contour $\gamma$ in $U$. Equivalently, $F$ has a complex antiderivative on $U$. Star-shaped means that there is one point of $U$ from which every straight segment to another point of $U$ remains in $U$.

Apply the Cauchy integral theorem to $F(z)=f(z)(L(z)-i\pi )$ on the slit plane, which is star-shaped about $-1$. For fixed $\epsilon >0$, first let $\delta \to 0$. The upper bank is traversed from $\epsilon$ to one and the lower bank from one to $\epsilon$, so their limiting sum isThe outer arc tends to the positively oriented unit-circle integral, interpreting the logarithm by its one-sided boundary values at the cut. On the inner arc, $|L-i\pi |\le |\log \epsilon |+\pi$ and $f$ is bounded near zero. Its integral is therefore $O(\epsilon |\log \epsilon |)$ and tends to zero. The sum of all four contour pieces is zero. Letting $\epsilon \to 0$ givesThe integral around the full circle is a piecewise integral of the branch's boundary values; $L$ is not a single holomorphic function across the cut-crossing point. This is the centred logarithmic keyhole identity.

Along $z=e^{i\theta }$ with $0<\theta <2\pi$, the chosen branch has $L(z)-i\pi =i(\theta -\pi )$. Use part (iii) and the circular estimate in part (a):The second step uses $|\theta -\pi |\le \pi$. The single branch-cut point does not affect the integral. Keeping that weight gives the sharper intermediate bound.

On the real interval, $h(x)$ is real and $H(x)=h(x{)}^2\ge 0$. Expanding and integrating the finite sum yieldsApply part (b)(iv) to the entire function $H=h^2$. Since $|H(e^{i\theta })|=|h(e^{i\theta }){|}^2$, orthogonality of complex exponentials givesConsequentlyThe quadratic form is actually nonnegative, so its absolute value can be dropped. This shifted Hilbert quadratic-form bound comes from combining the logarithmic contour estimate with a finite Parseval identity.