 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-1/2a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

At the origin, the Taylor series

sin z = z + O (z^{3})

and

1 - cos z = z^{2} /2 + O (z^{4})

give

f_{1} (z) = z^{- 3} (1 + O (z)), f_{2} (z) = 2 z^{- 4} (1 + O (z^{2})) .

Thus $f_{1}$ has a pole of order three and $f_{2}$ a pole of order four at zero. For the third function, the Laurent series

f_{3} (z) = \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k} z ^{1 - 2 k}}{( 2 k + 1 )!}

has infinitely many nonzero negative-power terms, so zero is an essential singularity of $f_{3}$ .

To classify the point at infinity on the Riemann sphere, use the coordinate

w = 1/ z

. Then

f_{3} (1/ w) = w^{- 2} sin w = w^{- 1} - \frac{w}{6} + O (w^{3}),

so $f_{3}$ has a simple pole at infinity.

For the other two functions, an important qualification is necessary. The poles of

f_{1}

at nonzero integer multiples of

π

, and those of

f_{2}

at nonzero integer multiples of

2 π

, approach infinity. Equivalently, their images in the

w

coordinate accumulate at zero. Hence infinity is a non-isolated singularity of $f_{1}$ and $f_{2}$ . Neither function is holomorphic on any punctured neighbourhood of infinity, so the classification of isolated singularities into removable singularities, poles and essential singularities does not apply there.

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