 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2006/analysis/4/ii/solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2006 / Analysis / 4 / ii / Solution by

Codex  0 2026-10-31

Write

r = x^{2} + y^{2} > 0

. For

u = x / (1 + r)

,

v = y / (1 + r)

,

u_{x} = \frac{1}{1 + r} - \frac{x ^{2}}{r ( 1 + r ) ^{2}}, v_{y} = \frac{1}{1 + r} - \frac{y ^{2}}{r ( 1 + r ) ^{2}}, u_{y} = v_{x} = - \frac{x y}{r ( 1 + r ) ^{2}} .

The two Cauchy-Riemann equations would require

x^{2} = y^{2}

and

x y = 0

simultaneously, impossible when

r > 0

. Thus

f

is not complex differentiable at any nonzero point.

At zero, however,

\frac{f ( h ) - f ( 0 )}{h} = \frac{1}{1 + ∣ h ∣} ⟶ 1.

So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $C$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.

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