Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 7

Write $z=x+iy$ and square the modulus equation. It becomesFor $P\ne 1$, completing the square gives the Apollonius circleHenceThe printed statement needs the exception $\boxed{P=1⟹y=0}$: equal distances from $ib$ and $-ib$ give their perpendicular bisector, a straight line rather than a finite-radius circle. For $P>1$ the center lies in the upper half-plane; this is the case needed in part (ii). The original PDF confirms the numerator $z+ib$, correcting the duplicated, corrupted OCR formula.

For the upper Apollonius circle, let its constant ratio be $P>1$. Matching the center and radius from part (i) givesSubtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. AlsoThe relevant inverse hyperbolic cosine is positive. Now $\log |(z+ib)/(z-ib)|$ is the real part of a local holomorphic logarithm, hence a harmonic function away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.

On $y=0$ the numerator and denominator have equal modulus, so the logarithm is zero. On the circular boundary its value is $\log P=\mathrm{arcosh}(y_0/R)$. Therefore the harmonic potential between a grounded line and a circular boundary isThis verifies Laplace's equation and both prescribed Dirichlet boundary conditions. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.

Substituting $z=1$ into the defining Gamma integral givesBoth endpoints are convergent, so no continuation of the Gamma function is needed here.

For real $z>0$, apply integration by parts to the Gamma integral with $u=t^z$ and $dv=e^{-t}dt$:The boundary term vanishes: $t^z\to 0$ at zero, and exponential decay dominates the power at infinity. This proves the Gamma function recurrence

Iterating the Gamma function recurrence gives a meromorphic continuationNear $z=-m$, the numerator is analytic and tends to $\Gamma (1)=1$. Exactly one denominator factor, $z+m$, vanishes, while all other factors tend to the nonzero product $(-1{)}^{m}m!$. ThereforeThus $\boxed{z=0,-1,-2,\dots \text{ are all simple poles}}$. More precisely, the Gamma function residue at a nonpositive integer is $\boxed{{\mathrm{Res}}_{z=-m}\Gamma (z)=(-1{)}^m/m!}$. Taking $m$ large enough to put $z+m+1$ in the original right half-plane also shows that this continuation introduces no other finite poles.