Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 3 / iii

Table of contents
- a iii
  - Solution a
- b iii
  - Solution b

a

Solution

The Cauchy-Riemann equations give

v_{x} = 2 x

,

v_{y} = 2 - 2 y

. Integrating and matching both derivatives gives the harmonic conjugate

v = x^{2} + 2 y - y^{2} + C, f (z) = 2 z + i z^{2} + i C, C \in R .

The real part of this polynomial is exactly the required

2 x (1 - y)

.

b

Solution

Here

v_{x} = - 3 x^{2} + 3 y^{2} + 4 y

and

v_{y} = 6 x y + 4 x

. The harmonic conjugate and resulting polynomial are

v = - x^{3} + 3 x y^{2} + 4 x y + C, f (z) = 2 z^{2} - i z^{3} + i C, C \in R .

Indeed

Re (- i z^{3}) = 3 x^{2} y - y^{3}

and

Re (2 z^{2}) = 2 x^{2} - 2 y^{2}

.

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