 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/5/b/iii/solution

A pole of order

n \geq 1

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

Equivalently, the Laurent series has a finite nonzero principal part, with lowest power exactly

(z - z_{0})^{- n}

. Thus

(z - z_{0})^{n} f (z)

extends holomorphically with nonzero value at

z_{0}

; using a smaller positive power would not remove the singularity.

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