Mean Value Theorem for Holomorphic Functions
Jump to navigation
Jump to search
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac i {2 \pi i} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \paren {r e^{i \theta} \times \frac 1 {r e^{i \theta} } } \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta\) |
$\blacksquare$