Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 4

A sufficient contour setting for both results is a positively oriented, simple, closed, piecewise smooth contour $C$, with the function holomorphic on an open neighbourhood of $C$ and its entire interior. The Cauchy integral theorem then statesIf $z_0$ is strictly inside $C$, the Cauchy integral formula statesThe contour must not pass through $z_0$. Positive orientation means anticlockwise; reversing it reverses the sign of the contour integral. The holomorphy assumption throughout the interior excludes hidden singularities there.

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore