 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/4/d/solution

For a double pole at

z_{0} \in {i, - i}

, differentiate the remaining holomorphic factor:

Res (f, z_{0}) = \frac{d}{d z} \frac{1 + z + z ^{2}}{( z + z _{0} ) ^{2} z}_{z = z_{0}} .

Since

z_{0}^{2} = - 1

and

1 + z_{0} + z_{0}^{2} = z_{0}

, this simplifies to

Res (f, z_{0}) = \frac{1/8 - z _{0} /2}{z _{0}} .

On the selected branch

i = e^{iπ /4}

and

- i = e^{3 iπ /4}

Res (f, i) = \frac{2}{16} (- 3 - 5 i), Res (f, - i) = \frac{2}{16} (3 - 5 i) .

The residue theorem therefore gives

C = 2 πi (- \frac{5 i 2}{8}) = \frac{5 π 2}{4}, I = \frac{5 π 2}{8} .

The equality

C = 2 I

uses the limiting contour integrals established in (c).

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