Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 12 (Complex Analysis OR Complex Methods)

For any circle $|\zeta |=\rho$ with $r<\rho <R$, Cauchy's formula applied inside and outside the circle gives the locally uniformly convergent Laurent expansionAt zero, the singularity is removable exactly when all $a_n$ with $n<0$ vanish; it is a pole of order $k$ when $a_{-k}\ne 0$ and $a_n=0$ for $n<-k$; it is essential when infinitely many negative coefficients are nonzero.

Since ${\csc }^{2}z=z^{-2}+1/3+O(z^2)$, the principal parts cancel and $f_1=-1/3+O(z^2)$. The singularity is removable.

Termwise integration of the exponential series givesThere are infinitely many negative Laurent coefficients, so zero is an essential singularity.

For distinct $z,w$ in the convex disc, integrate along the segment:The second term has modulus strictly below $|f^{\prime }(z_0)||z-w|$, so the sum cannot vanish. Thus $f$ is one-to-one.

Harmonic means twice continuously differentiable with $u_{xx}+u_{yy}=0$. On the simply connected plane, choose an entire harmonic conjugate so $F=u+iv$ is entire. If $u\ge 0$, then $e^{-F}$ is bounded; Liouville's theorem makes it, and hence $u$, constant.

Choose an entire $F$ with real part $u$. The bound implies that $e^F$ has at most polynomial growth, so Cauchy's estimates make it a polynomial. It has no zeros, hence that polynomial is constant. Therefore $F$, and in particular $u$, is constant.