Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 4

A function is an analytic function at $z_0$ if it is holomorphic throughout some open neighbourhood of $z_0$, equivalently if it has a convergent power series there. In real components $g=u+iv$, continuous first partial derivatives in a neighbourhood together with the Cauchy-Riemann equationsthroughout that neighbourhood are sufficient, and for an analytic function they hold there. Satisfying these equations only at the single point is not enough to establish analyticity in a neighbourhood.

Factor the polynomial denominator as $(z-i{)}^2(z+i{)}^2$. The numerator has values $i$ and $-i$ at the two roots, so neither cancels. ThereforeThe factor $z^{-1/2}$ gives a branch point at zero, with local behaviour $f(z)\sim z^{-1/2}$. Infinity is also a branch point, since in the coordinate $w=1/z$ the leading behaviour is $f(1/w)\sim w^{5/2}$. A single-valued branch requires a branch cut joining these points. Away from that cut and the two poles, the selected branch is holomorphic. For the integral below, use $0<\arg z<2\pi$, putting the cut along the positive real axis.

Use a counterclockwise keyhole contour around the positive real branch cut, with small radius $\epsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\epsilon$, so its contribution also tends to $I$.

The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O({\epsilon }^{1/2})$, since $f=O({\epsilon }^{-1/2})$. Both vanish. HenceThe contour encloses both double poles $\pm i$ but excludes the branch point.

For a double pole at $z_0\in \{i,-i\}$, differentiate the remaining holomorphic factor:Since $z_0^2=-1$ and $1+z_0+z_0^2=z_0$, this simplifies toOn the selected branch $\sqrt{i}=e^{i\pi /4}$ and $\sqrt{-i}=e^{3i\pi /4}$, givingThe residue theorem therefore givesThe equality $C=2I$ uses the limiting contour integrals established in (c).