Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 4

Write $f=u+iv$ and $z=x+iy$. For a positively oriented, piecewise smooth simple closed contour bounding a region where $u,v$ have continuous first partial derivatives, Green's theorem givesThe Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$, so both integrands vanish:This proves Cauchy's integral theorem under the standard regularity assumptions for Green's theorem. More general admissible contours follow by subdivision; every enclosed region must be free of singularities.

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

The exponential numerator is entire and never zero. The denominator vanishes precisely when $e^{2z}=-1$, givingAt these points $\sinh z_n=i(-1{)}^n\ne 0$, so every zero of the denominator is simple and hence every singularity is a simple pole. Using the residue formula in (b),

Use the counterclockwise rectangular contour with vertices $-R,R,R+i\pi ,-R+i\pi$. It encloses just the simple pole $i\pi /2$, whose residue is $e^{k\pi /2}/i$. Let $I_R={\int }_{-R}^R{e}^{-ikx}/\cosh xdx$. The identitymakes the top edge, traversed right to left, contribute $e^{k\pi }{I}_R$. On either vertical edge,Thus the sum of the vertical integrals tends to zero, with absolute value at most $2\pi e^{|k|\pi }/\sinh R$. The residue theorem and $R\to \infty$ now giveThe argument works for every fixed real $k$, including zero and negative values, and gives the Fourier transform of the hyperbolic secant with the convention $\tilde{f}(k)=\int e^{-ikx}f(x)dx$.