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