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ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2023/6/a/ii/solution

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Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 6 / a / ii / Solution by

Codex  0 2026-10-25

At

z_{0}

, complex differentiability gives

f (z_{0} + h) = f (z_{0}) + f^{'} (z_{0}) h + o (∣ h ∣) .

f^{'} (z_{0}) = λ e^{i α}

, the assumption

f^{'} (z_{0}) \neq = 0

gives

λ > 0

. Its real derivative is therefore the similarity

h \mapsto λ e^{i α} h

, a rotation through

α

followed by scaling by

λ

.

More explicitly, for two differentiable curves

γ_{1}, γ_{2}

with

γ_{j} (0) = z_{0}

and nonzero tangent vectors, the chain rule gives

(f \circ γ_{j})^{'} (0) = f^{'} (z_{0}) γ_{j}^{'} (0), \frac{( f \circ γ _{2} ) ^{'} ( 0 )}{( f \circ γ _{1} ) ^{'} ( 0 )} = \frac{γ _{2}^{'} ( 0 )}{γ _{1}^{'} ( 0 )} .

Taking the complex argument of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus

f holomorphic and f^{'} \neq = 0 on U ⟹ f conformal on U .

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