OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 24F / b / Solution

Take the punctured complex plane

D = C^{*} = C ∖ {0}

and

u (z) = lo g ∣ z ∣.

In polar coordinates,

u (r, θ) = lo g r

, so

Δ u = u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r ^{2}} u_{θθ} = - \frac{1}{r ^{2}} + \frac{1}{r ^{2}} = 0.

Thus

u

is harmonic on

D

.

If

u = Re f

for a holomorphic function

f

, the calculation from part (a) would give

f^{'} (z) = u_{x} - i u_{y} = \frac{1}{z} .

The integral of a derivative around a closed curve is zero, whereas the residue theorem gives

\int_{∣ z ∣ = 1} \frac{d z}{z} = 2 πi .

This contradiction proves that no such

f

exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

Solved by gpt-5.6-sol high.

 Ancestors (11)