Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 7D (Further Complex Methods)

For any $n$ with $\mathrm{Re}z>-n$, defineThe definitions agree on overlaps by the recurrence, giving a meromorphic continuation to $\mathbb{C}$. Its only singularities are simple poles at $0,-1,-2,\dots$. Since $F(1)=1$, repeated use of the recurrence givesThese residues are nonzero, so all the listed singularities really are simple poles.

The meromorphic continuations of $F$ and the gamma function have the same recurrence and the same value at $1$. Hence their residues at every nonpositive integer are the same. Their difference $f=F-\Gamma$ therefore has removable singularities there and is analytic elsewhere, so it extends to an entire function. Equivalently, $f(z+1)=zf(z)$ and $f(1)=0$ show inductively that the apparent quotient singularities are removable.

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,Its modulus is therefore periodic with period one. The assumed bound for $f$ on $0\le \mathrm{Re}z\le 1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb{C}$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an open set; the identity theorem would then give $f\equiv 0$, a contradiction to that choice. Therefore $f\equiv 0$ and $F=\Gamma$.