 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/analysis/6/a/iv/solution

A zero of order

n \geq 1

a

means that

f

is holomorphic near

a

and

f (a) = f^{'} (a) = \dots = f^{(n - 1)} (a) = 0, f^{(n)} (a) \neq = 0.

Equivalently its Taylor series starts with the nonzero coefficient of

(z - a)^{n}

, so

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

Unlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

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