Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S01 (Functions of a complex variable)

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.

Conformality and analyticity. Use the orientation-preserving convention for a conformal map, with a continuously differentiable, nonzero differential. Its Jacobian matrix must be a positive scale times a rotation,Thus $u_x=v_y$, $u_y=-v_x$, the Cauchy-Riemann equations; continuous partial derivatives then imply that $u+iv$ is a holomorphic function. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto \bar{z}$ preserves unsigned angles but is an anticonformal map and is not a holomorphic function.

The mapping. Put $a=\pi /L$. The displayed components combine to giveIts derivative does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a conformal map there. Also $\mathrm{Im}w=\sinh (ax)\sin (ay)>0$. The lower edge maps to $[1,\infty )$, the upper edge to $(-\infty ,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the upper half-plane. Indeed, $\cosh \zeta =\cosh \eta$ implies $\zeta =\pm \eta +2\pi im$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of conformality.

Transformed boundary values. A conformal map preserves Laplace's equation up to a positive scale. Thus the transformed Dirichlet problem is to find a real harmonic function $U(u,v)$ on $v>0$ withThe conflicting values at the original corners are interpreted as one-sided edge limits.

Logarithmic solution. Choose the complex logarithms of $w\pm 1$ with arguments between zero and $\pi$ in the upper half-plane. A linear combination $F(w)=\alpha \log (w+1)+\beta \log (w-1)+\gamma$ is a complex-valued harmonic function, since it is a holomorphic function; the real physical potential must be a real or imaginary part. In particular, setThenOn the three real boundary intervals, the argument difference is respectively $0,\pi ,0$, which gives the required values. This is the bounded harmonic solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the maximum principle. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh (\pi x/L)\sin (\pi y/L)$.

Back substitution. Sincethe required continuous argument difference is $2\arctan [\sin (ay)/\sinh (ax)]$ in the open half-strip. ConsequentlyAs $y\to 0,L$ with $x>0$, the arctangent tends to zero; as $x\to 0^{+}$ with $0<y<L$, it tends to $\pi /2$. These limits verify every open edge of the Dirichlet problem.

Singularities and cuts. There are four simple poles $z_j=e^{i(\pi /4+j\pi /2)}$, $j=0,1,2,3$, because the roots of $1+z^4$ are simple and do not coincide with $\pm 1$. At $z=\pm 1$, the local behavior is a nonzero constant times $(z\mp 1{)}^{-1/2}$: these are square-root branch points. Infinity is not a branch point; on either branch $f(z)=O(z^{-5})$, so it extends there with a zero of order five. Three valid choices of branch cuts are the segment $[-1,1]$, the two rays $(-\infty ,-1]\cup [1,\infty )$ joined through infinity, or a semicircular arc joining $-1$ to $1$ and avoiding the poles. A circuit about either endpoint changes the sign, while a circuit about both endpoints does not.

Branch and residues. On $\mathbb{C}\setminus [-1,1]$, let $s(z)=\sqrt{z^2-1}$ be the branch with $s(z)/z\to 1$ at infinity, and define $\sqrt{1-z^2}=-is(z)$. Its upper boundary value on $(-1,1)$ is positive. The residue at a simple pole $z_j$ isFor fully explicit values, putIn counterclockwise order $z_0,z_1,z_2,z_3$, the values of $s$ are $A+iB,-A+iB,-A-iB,A-iB$. Since $A^2+B^2=\sqrt{2}$, substitution yieldsA branch point has no ordinary residue, since it has no single-valued Laurent series on a punctured neighborhood. At infinity the residue is zero.

Contour evaluation. The original contour goes counterclockwise around a large circle and clockwise round the slit, with its upper lip directed to the right. The large-circle contribution is $O(R^{-4})$. Small endpoint arcs contribute $O({ϵ}^{1/2})$. The artificial slit extension from $1$ to $R$ has equal boundary values on both sides, so its contributions cancel. On $(-1,1)$ the two boundary values of $f$ are opposite. Thus, with $I$ denoting the requested real integral, the residue theorem givesTherefore