Past exam of the metric spaces and complex analysis course of the University of Oxford / 2015 / 4

Choose $0<\rho <r$ and traverse the circle $|z|=\rho$ counterclockwise, so it has positive orientation of a contour. The Cauchy coefficient formula gives, for every nonnegative integer $n$,The smaller radius matters: the holomorphic function is only known on $|z|<r$, so its values on $|z|=r$ need not exist. To see directly why this contour integral selects $c_n$, the power series converges uniformly on $|z|=\rho$, allowing uniform convergence and contour integration to justify integrating term by term. The contour integral of $z^{k-n-1}$ is zero for $k\ne n$ and $2\pi i$ for $k=n$. Thus all other terms disappear.

Use the generalized binomial theorem with exponent $-1/2$, choosing the analytic branch of a square root that equals $1$ at $z=0$. For $|z|<1$,The signs cancel in each binomial coefficient. For $n\ge 1$,The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the power series isIts coefficients are the normalized central binomial coefficients. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the ratio test gives absolute convergence inside the unit disc. Applying the Cauchy coefficient formula to the holomorphic function $(1-z{)}^{-1/2}$ now immediately yieldsThis value is independent of the chosen radius because the contour integral extracts the same coefficient on every such circle.

For $z\in \mathbb{C}\setminus [1,\infty )$, the number $w=1-z$ avoids the nonpositive real axis. Define the branch of the complex logarithm byHere the principal argument is selected as a continuous map on that slit domain, so this is a holomorphic function, with derivative $-1/(1-z)$ and $L(1)=0$. For example, locally writing $w=\rho e^{i\varphi }$ gives $d(\log \rho +i\varphi )=dw/w$, verifying the complex derivative. The normalization also makes this branch of the complex logarithm unique: any two continuous choices differ by a constant integer multiple of $2\pi i$ on the connected slit domain, and their values at $w=1$ agree.

To evaluate the corresponding analytic branch of a square root on the unit circle, factorFor $0<\theta <2\pi$, we have $\sin (\theta /2)>0$ and $-\pi /2<\theta /2-\pi /2<\pi /2$. Thus this factorization already uses the selected principal argument, with no hidden change by $2\pi$. It follows thatHalving this expression and applying the exponential function givesSince $\sqrt{2}{e}^{-i\pi /4}=1-i$, the desired branch value isFor instance, $\theta =\pi$ gives $\sqrt{2}>0$, a quick check that the sign agrees with the normalization.

Write $b_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The given limiting identity and the Cauchy coefficient formula from part (a) give the boundary contour integralThe contour passes through $z=1$, so its value here is an improper integral using the one-sided branch values. Parametrize its open arc by $z=e^{i\theta }$, $0<\theta <2\pi$. Substitution of the analytic branch of a square root calculated in part (b) givesThe factor $e^{-in\theta }$ comes from cancelling $e^{i\theta }$ in $dz$ against $z^{n+1}$; the extra $e^{-i\theta /4}$ comes from the chosen branch. Keeping both phases prevents a change in the half-integer frequency.

Now put $t=\theta /2$. ThenDividing by $i-1$ givesTaking the real part therefore provesAll these improper integrals converge absolutely: the numerators are bounded, while $(\sin t{)}^{-1/2}$ is of order $t^{-1/2}$ at zero and $(\pi -t{)}^{-1/2}$ at $\pi$. Hence the substitutions and passage to the real part are legitimate. The imaginary part also shows that the corresponding sine integral has the same positive value, which checks the phase calculation.