MATH3401

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

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 \(C_{\rho}\).

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

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 \[ |f(z)-f(z_0)| \lt \varepsilon \quad \forall |z-z_0|\lt \delta \quad (3) \]

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