 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/4/i/solution

For a counterclockwise simple contour containing

z_{0}

, and

f

analytic on a neighbourhood of its interior and boundary, the Cauchy integral formula is

\frac{1}{2 πi} \oint_{C} \frac{f ( z )}{z - z _{0}} d z = f (z_{0}) .

\frac{1}{z ^{2} - 1} = \frac{1}{2} (\frac{1}{z - 1} - \frac{1}{z + 1}) .

Both points

1, - 1

lie inside

∣ z ∣ = 2

. Applying the formula with the constant analytic function one gives

2 πi

for each of the two terms. They cancel, so

\oint_{∣ z ∣ = 2} \frac{d z}{z ^{2} - 1} = \frac{1}{2} (2 πi - 2 πi) = 0.

Reversing the circle's orientation would leave this zero result unchanged.

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