 Solution

ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2007/b/6/a/solution

f^{'} (0) \neq = 0

, so

a_{1} \neq = 0

. Apply the stated Green's theorem with

F = z^{n} \overset{z}{ˉ}

M_{n} = \frac{1}{2 i} \oint_{Γ} z^{n} \overset{z}{ˉ} d z = \frac{1}{2 i} \oint_{∣ ζ ∣ = 1} f (ζ)^{n} (\sum_{j = 1}^{N} \overset{a}{ˉ}_{j} ζ^{- j}) f^{'} (ζ) d ζ .

Here

\overset{ˉ}{ζ} = ζ^{- 1}

on the unit circle. The lowest power in

f^{n} f^{'}

ζ^{n}

, while the most negative power in the conjugate polynomial is

ζ^{- N}

. For

n \geq N

their product has no

ζ^{- 1}

term. The residue theorem gives $M_{n} = 0$ for all $n \geq N$ .

When

N = 2

, the residue for

n = 0

∣ a_{1} ∣^{2} + 2∣ a_{2} ∣^{2}

. For

n = 1

, expand

f f^{'} = a_{1}^{2} ζ + 3 a_{1} a_{2} ζ^{2} + 2 a_{2}^{2} ζ^{3} .

Only

a_{1}^{2} ζ \overset{a}{ˉ}_{2} ζ^{- 2}

contributes. Therefore

M_{0} = π (∣ a_{1} ∣^{2} + 2∣ a_{2} ∣^{2}), M_{1} = π a_{1}^{2} \overset{a}{ˉ}_{2} .

The first is the area; the second records its complex first moment, and need not be real.

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