MATH3401

Lecture 21

Cauchy Integral formula

Let $$f$$ be analytic on and inside a simple closed curve $$C$$ that is oriented (i. e. if $$z(t)$$ parametrises $$C$$ then as $$t \nearrow$$, $$\text{Int}\, C$$ stays on the curve's LHS).

Then if $$z_0 \in \text{Int}\, C$$ we have

$$f(z_0) = \ds \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0} dz, \qquad (1)$$,

i. e.   $$2 \pi i \,f(z_0) = \ds \int_C \frac{f(z)}{z-z_0} dz. \quad (1)'$$

Cauchy Integral formula

Proof: Set $C_{\rho} = \left\{ z(\theta) = z_0 + \rho e^{i \theta}, 0\leq \theta \leq 2 \pi\right\},$ for $$\rho$$ sufficiently small such that $\text{Int}\, C_{\rho} \subset D = \text{Int}\, C.$

Then $$\ds \frac{f(z)}{z-z_0}$$ is analytic on $$\text{Int}\, C \setminus \text{Int}\, C_{\rho}$$ and $$C$$ and $$C_{\rho}$$.

Cauchy Integral formula

Proof (cont): C-G extension implies $\int_C \frac{f\left(z\right)dz}{z-z_0} = \int_{C_{\rho}} \frac{f\left(z\right)dz}{z-z_0}.$

Then $\small \int_C \frac{f\left(z\right)dz}{z-z_0} - f\left(z_0\right) \int_{C_{\rho}} \frac{dz}{z-z_0} = \int_{C_{\rho}} \frac{f\left(z\right) - f\left(z_0\right) }{z-z_0} dz \quad (2)$

Cauchy Integral formula

Proof (cont): But $\int_{C_{\rho}} \frac{dz}{z-z_0} = 2 \pi i$ (c.f. Lecture 20). Since $$f$$ is analytic at $$z_0$$, it is continuous at $$z_0$$.
Then, given $$\varepsilon \gt 0$$ exists $$\delta \gt 0$$ such that $|f(z)-f(z_0)| \lt \varepsilon \quad \forall |z-z_0|\lt \delta \quad (3)$

Cauchy Integral formula

Proof (cont): Choose $$\rho \lt \delta$$. Then $$\left| f\left( z_0\right)+\rho e^{i\theta} - f\left( z_0\right) \right| \lt \varepsilon$$    ðŸ˜ƒ

$$\overset{{\tiny(2)}}{\implies} \left|\ds \int_C \frac{f\left(z\right)dz}{z-z_0} - 2 \pi i\, f(z_0) \right| \le \ds \int_{C_{\rho}} \frac{\left|\, f\left(z\right) - f\left(z_0\right) \right| }{\left|z-z_0\right|} dz$$

$$\;\; \; = \ds \frac{1}{\rho} \int_{C_{\rho}} \big| \,f\left(z\right) - f\left(z_0\right) \big| dz \qquad \qquad$$

$$\quad \qquad\lt \ds \frac{1}{\rho} \cdot \varepsilon \cdot 2 \pi \rho \qquad$$ (via ðŸ˜ƒ and $$M$$-$$\ell$$)

$$= 2 \pi \varepsilon.$$   Send $$\varepsilon \searrow 0 \implies$$ $$(1)'$$. $$\square$$