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

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.