Here is a neat upper bound on the derivative of a complex function. This is one of those things that made me wish I studied complex analysis. $\newcommand{\abs}[1]{| #1 |} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}}$

Lemma: Let $f : U \longrightarrow \C$ be holomorphic and suppose that $D_R = \{ z : \abs{z - z_0} \leq R \} \subset U$. Let $\gamma_R = \{ z : \abs{z - z_0} = R \}$ denote the boundary of $D_R$. For all $n \geq 0$, $$\abs{f^{(n)}(z_0)} \leq \frac{n!}{R^n} \max_{z \in \gamma_R} \abs{f(z)} \:.$$

Proof: This follows almost directly from Cauchy's integral formula, which states that $$f^{(n)}(z_0) = \frac{n!}{2\pi j} \oint_{\gamma_R} \frac{f(z)}{(z-z_0)^{n+1}} \; dz \:.$$ Hence, by estimating $\abs{f(z)}$ from above with the maximum on $\gamma_R$, we conclude that $$\begin{align*} \abs{f^{(n)}(z_0)} &\leq \frac{n!}{2\pi} \max_{z \in \gamma_R} \abs{f(z)} \int_{0}^{2\pi} \frac{R}{R^{n+1}} \; d\theta = \frac{n!}{R^n} \max_{z \in \gamma_R} \abs{f(z)} \:. \end{align*}$$ $\square$