Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 13A (Complex Analysis or Complex Methods)

Fix $\mathrm{Im}\lambda >-c$. Truncate $C_1$ at $\pm R-ic$ and close it with the upper semicircle centred at $-ic$, oriented counterclockwise. For large $R$ the only pole of the integrand inside is $z=\lambda$. Uniformly on the arc, $|z-\lambda |$ is comparable to $R$ and $f(z)=o(R^{-1})$, so the arc integral is $o(R^{-1})$ and tends to zero. The line integral is absolutely convergent at infinity for the same decay reason. The Cauchy integral formula givesThe right-hand side is holomorphic on this open half-plane by the hypothesis, proving the required holomorphic function. The contour never passes through $\lambda$ because its imaginary part is strictly above the line.

Write $r_{+}={\mathrm{Res}}_{z={\alpha }_{+}}g(z)$. For $\mathrm{Im}\lambda >-c$ and $\lambda \ne {\alpha }_{+}$, close $C_1$ upwards as before. The residue theorem includes the poles at $\lambda$ and ${\alpha }_{+}$, and the assumed decay in the upper region makes the arc contribution vanish. ThusIt is essential to include the apparently exceptional point $\lambda ={\alpha }_{+}$. The Laurent series there is $g(z)=r_{+}/(z-{\alpha }_{+})+g_0(z)$ with $g_0$ holomorphic. At that point the integrand is $r_{+}/(z-{\alpha }_{+}{)}^2+g_0(z)/(z-{\alpha }_{+})$, whose residue is $g_0({\alpha }_{+})$. HenceThe displayed difference is interpreted by its removable singularity. It therefore defines a holomorphic function at every point of the required half-plane, including ${\alpha }_{+}$.

For $\mathrm{Im}\lambda <c$, close $C_2$ upwards. The pole at $\lambda$ is outside this contour, while ${\alpha }_{+}$ is inside; the pole ${\alpha }_{-}$ is also outside. The residue theorem and the minus sign in the definition of $J$ giveThis rational function is holomorphic throughout $\mathrm{Im}\lambda <c$, since its only pole lies above $c$. In particular the formula includes $\lambda ={\alpha }_{-}$: the original line-integrand has no singularity on $C_2$ there, and no evaluation of $g({\alpha }_{-})$ is needed. Closing downwards would require a decay hypothesis in the lower half-plane that has not been given; closing upwards avoids that unjustified step.

In the open strip $-c<\mathrm{Im}\lambda <c$, neither pole of $g$ is present. The two evaluations therefore giveBoth summands are holomorphic on this strip. This is a half-plane decomposition by Cauchy integrals; the pole subtraction in $H$ is what makes its extension holomorphic even at the upper pole of $g$.