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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/4/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 4 / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Compute the complex derivative along horizontal and vertical increments. Along the real direction,

f^{'} = u_{x} + i v_{x}

; along the imaginary direction,

f^{'} = (u_{y} + i v_{y}) / i = v_{y} - i u_{y}

. Equality gives the Cauchy-Riemann equations

u_{x} = v_{y}, u_{y} = - v_{x} .

The real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives

u_{xx} + u_{yy} = v_{y x} - v_{x y} = 0

and

v_{xx} + v_{yy} = - u_{y x} + u_{x y} = 0

, proving that both are harmonic functions.

For the first real part,

u_{x} = 2 x

and

u_{y} = - 2 y

. The Cauchy-Riemann equations require

v_{y} = 2 x

and

v_{x} = 2 y

, so its harmonic conjugate is

v = 2 x y + C

. Therefore

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

For the second real part,

u = x / (x^{2} + y^{2})

is the real part of

1/ z

. Its harmonic conjugate is

- y / (x^{2} + y^{2}) + C

f (z) = \frac{1}{z} + i C, C \in R, z \neq = 0.

These are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.

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