Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 4

For a counterclockwise simple contour containing $z_0$, and $f$ analytic on a neighbourhood of its interior and boundary, the Cauchy integral formula isFor the required rational integral, use partial fraction decomposition:Both points $1,-1$ lie inside $|z|=2$. Applying the formula with the constant analytic function one gives $2\pi i$ for each of the two terms. They cancel, soReversing the circle's orientation would leave this zero result unchanged.

On a positively oriented contour encircling $z_0$ once inside an annulus of analyticity, the Laurent series coefficients areThis follows by inserting the convergent Laurent expansion and integrating term by term: only the term with exponent $-1$ has nonzero contour integral. Uniform convergence on compact circles in the annulus justifies that step.

For the function at hand,The geometric series converges for $|z|<1$, and the pole at zero excludes the centre. ThusThe coefficients are $a_n=-1$ for $n\ge -1$ and zero for $n<-1$. The coefficient formula gives the same result on $|z|=\rho <1$: $a_n=(2\pi i{)}^{-1}\oint z^{-n-2}/(z-1)dz$, and expansion of $(z-1{)}^{-1}$ selects exactly those coefficients.

The annulus is essential. For $|z|>1$, expand instead in inverse powers:These are different Laurent expansions about the same centre on the two annuli separated by the singularity at $z=1$. The series requested in the PDF is the inner one, not a global expansion.