 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/4/a/ii/solution

The denominator vanishes at the four roots of

z^{4} = - 1

, namely

z_{k} = e^{i (2 k + 1) π /4}

for

k = 0, 1, 2, 3

. They are simple poles, since

(1 + z^{4})^{'} = 4 z^{3}

is nonzero at each root and the numerator does not vanish there. The simple-pole residue formula gives

Res_{z = z_{k}} \frac{z ^{2}}{1 + z ^{4}} = \frac{z _{k}^{2}}{4 z _{k}^{3}} = \frac{1}{4 z _{k}} .

Explicitly,

pole (1 + i) / 2 (- 1 + i) / 2 (- 1 - i) / 2 (1 - i) / 2 residue (1 - i) / (42) (- 1 - i) / (42) (- 1 + i) / (42) (1 + i) / (42) .

There are no branch points or other finite singularities. At infinity the function is

O (z^{- 2})

, so infinity is removable rather than a pole.

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