Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 5 / ii

Table of contents
- a ii
  - Solution a
- b ii
  - Solution b

a

Solution

An entire antiderivative is

F (z) = z sin z + cos z

, since differentiation cancels the sine terms and leaves

F^{'} (z) = z cos z

. The curve runs from

- i

to

i

. Both terms of

F

are even, so

F (i) = F (- i)

, and the complex line integral gives

\int_{γ} z cos z d z = 0.

b

Solution

Along the circle,

z = z_{0} + r e^{i t}

and

d z = i r e^{i t} d t

. Expanding the nonholomorphic integrand gives

∣ z ∣^{2} = ∣ z_{0} ∣^{2} + r^{2} + r \overline{z}_{0} e^{i t} + r z_{0} e^{- i t} .

After multiplication by

i r e^{i t}

, all nonconstant integer-frequency exponentials have zero integral over

[0, 2 π]

. The remaining constant term is

i r^{2} z_{0}

. Therefore

\int_{γ} ∣ z ∣^{2} d z = 2 πi r^{2} z_{0} .

This direct parametrization avoids applying a holomorphic-antiderivative argument to

∣ z ∣^{2}

, which is not holomorphic.

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