 Solution

ID: past-exam-of-the-algebra-and-analysis-i-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/analysis/6/a/iii/solution

A pole of order

m \geq 1

a

has finitely many negative powers in its Laurent series, the lowest being exactly

- m

c_{- m} \neq = 0, c_{j} = 0 (j < - m) .

Equivalently,

f (z) = (z - a)^{- m} h (z), h holomorphic near a, h (a) \neq = 0.

Thus

(z - a)^{m} f (z)

a

, whereas multiplying by any smaller power does not remove the singularity.

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