Linear Bound between Complex Function and Derivative
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Theorem
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.
Let $z_0 \in D$.
Let $\epsilon \in \R_{>0}$.
Then there exists $r \in \R_{>0}$ such that for all $z \in \map {B_r} {z_0}$:
- $\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
where $\map {B_r} {z_0}$ denotes the open ball with center $z_0$ and radius $r$.
Proof
\(\ds \map {f'} {z_0}\) | \(=\) | \(\ds \ds \lim_{z \mathop \to z_0} \dfrac{ \map f z - \map f {z_0} } {z - z_0}\) | Definition of Complex-Differentiable at Point | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds \lim_{z_0 \mathop \to z} \dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren{ z - z_0 } } {z - z_0}\) | Combination Theorem for Limits of Complex Functions | ||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z_0 \mathop \to z} \size {\dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } {z - z_0} }\) | Modulus of Limit | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z_0 \mathop \to z} \dfrac {\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } } {\size {z - z_0} }\) | Complex Modulus of Quotient of Complex Numbers |
Given $\epsilon > 0$, we can find $r > 0$ by definition of limit such that for all $z \in \map {B_r} {z_0}$:
- $\dfrac {\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } } {\size {z - z_0} } < \epsilon$
We rearrange to:
- $\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori $\S 3.1$