Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 23F / Solution

Uniqueness in the defining characterization gives $\wp (-z)=\wp (z)$: the reflected function has the same poles, periodicity and prescribed principal part. Hence ${\wp }^{\prime }$ is odd and periodic. At each nonzero half-period $\omega \in \{1/2,\tau /2,(1+\tau )/2\}$, $-\omega \equiv \omega (\mathrm{mod}\Lambda )$, so ${\wp }^{\prime }(\omega )=-{\wp }^{\prime }(\omega )=0$.

An elliptic function has equally many zeros and poles, counted with multiplicities, in a fundamental parallelogram. This standard property applies to ${\wp }^{\prime }$, whose sole pole modulo the lattice has order three. The three distinct nonzero half-periods already supply three zeros, so these are all its zeros and each is simple. ThusEvenness and the prescribed expansion give $\wp (z)=z^{-2}+a_2{z}^2+a_4{z}^4+O(z^6)$. There is no constant term because the defining difference tends to zero, and only even powers occur. ConsequentlyAdding $20a_2\wp$ cancels the last pole. The resulting elliptic function is holomorphic everywhere and therefore constant, by boundedness on a fundamental parallelogram and Liouville theorem. Its value at zero is $-28a_4$. ThusAt the three half-periods $Q(\wp (\omega ))=0$. Their values are distinct: if two such points had the same value $e$, then $\wp -e$ would have a zero of multiplicity at least two at each, because its derivative vanishes there. That gives at least four zeros, contrary to its sole double pole and the zero-pole counting property. Hence the cubic has three distinct roots and leading coefficient four, provingAll properties specific to $\wp$ used here were derived from its defining characterization; only the general elliptic zero-pole count and holomorphic constancy were used as standard facts.