Mean Value Theorem for Holomorphic Functions

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Theorem

Let $D$ be a region.

Let $f : D \to \C$ be a holomorphic function.

Let $z \in D$.

Let $r$ be such that $\map {B_r} z \subseteq D$.


Then:

$\ds \map f z = \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$


Proof

By Cauchy's Integral Formula, we have:

$\ds \map f z = \frac 1 {2 \pi i} \oint_{\partial \map {B_r} z} \frac {\map f t} {t - z} \rd t$

where $\partial \map {B_r} z$ is the boundary of $\map {B_r} z$.

That is, $\partial \map {B_r} z$ is the circle of radius $r$, centred at $z$.

Note that we can parameterise $\partial \map {B_r} z$ by the function $\gamma : \hointr 0 {2 \pi} \to \partial \map {B_r} z$ defined by:

$\map \gamma \theta = z + r e^{i \theta}$

for each $\theta \in \hointr 0 {2 \pi}$.

Then, by the definition of a contour integral, we have:

\(\ds \frac 1 {2 \pi i} \oint_{\partial \map {B_r} z} \frac {\map f t} {t - z} \rd t\) \(=\) \(\ds \frac 1 {2 \pi i} \int_0^{2 \pi} \paren {i r e^{i \theta} \times \frac {\map f {z + r e^{i \theta} } } {z + r e^{i \theta} - z} } \rd \theta\) Derivative of Exponential Function
\(\ds \) \(=\) \(\ds \frac i {2 \pi i} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \paren {r e^{i \theta} \times \frac 1 {r e^{i \theta} } } \rd \theta\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta\)

$\blacksquare$