Spring 2011 - Problem 7

Solution.

Fix a triangle $\triangle$ with vertices $\set{z_1, z_2, z_3}$ oriented counter-clockwise. Suppose otherwise, so that there exists $\epsilon > 0$ such that

Decompose $\triangle_1 \coloneqq \triangle$ into four further triangles: $\triangle_1^{\p{1}}, \triangle_1^{\p{2}}, \triangle_1^{\p{3}}, \triangle_1^{\p{4}}$ so that

which have disjoint interiors and oriented counter-clockwise. By cancellation,

by the triangle inequality. Thus, there exists $j_0$ so that $\abs{\int_{\partial\triangle_1^{\p{j_0}}} f\p{z} \,\diff{z}} \geq \frac{\epsilon}{4}$ or else the triangle inequality implies

which is impossible. Set $\triangle_2 = \triangle_1^{\p{j_0}}$. By induction (e.g., by replacing $\triangle$ above with $\triangle_n$), we get a sequence of triangles $\set{\triangle_n}_n$ such that $\triangle_n \supseteq \triangle_{n+1}$ and $\abs{\int_{\partial\triangle_n} f\p{z} \,\diff{z}} \geq \frac{\epsilon}{4^{n-1}}$. Notice also that $\diam{\triangle_n} = \frac{\diam{\triangle_1}}{2^{n-1}} \to 0$ as $n \to \infty$, so by completeness of $\C$,

Since $f$ is complex differentiable, $f'\p{z^*}$ exists and if we set $h\p{z} = \frac{f\p{z} - f\p{z^*}}{z - z^*} - f'\p{z^*}$, then $h\p{z} \to 0$ as $z \to z^*$ and

Moreover, there exists $\delta > 0$ so that $\abs{h\p{z}} < \eta$, so because $\diam\triangle_n \to 0$, there exists $n \geq 1$ such that $\triangle_n \subseteq B\p{z^*, \delta}$. Observe that by the fundamental theorem of calculus, $\int_\gamma az + b \,\diff{z} = 0$ for any closed curve $\gamma$. Hence,

Letting $\eta \to 0$, however, we see that $\epsilon = 0$, which is impossible. Thus, $\epsilon$ must have been $0$ to begin with, which completes the proof.