Cauchy-Riemann
Let $f(z) = f(x + iy) = g(x, y) + ih(x, y)$. If it’s holomorphic, we can differentiate along the real line and the imaginary line:
$f’ = \lim\frac{f(x + a, y) – f(x, y)}{a} = \frac{∂g}{∂x} + i\frac{∂h}{∂x}$
$f’ = \lim\frac{f(x , y + a) – f(x, y)}{ia} = \frac{∂h}{∂y} – i\frac{∂g}{∂y}$
So we have $\frac{∂g}{∂x} = \frac{∂h}{∂y}$ and $\frac{∂h}{∂x} = -\frac{∂g}{∂y}$.
Power series
Theorem. Let $f(z) = \Sigma a_n z^n$. The radius of convergence is then given by $R = \frac{1}{\limsup \sqrt[n]{|a_n|}}$.
Theorem. Let $f(z) = \Sigma a_n z^n$ have radius of convergence R. Then f has a derivative $f'(z) = \Sigma n a_n z^{n-1}$ with identical radius of convergence.
Proof: First, set $g(z) = \sum na_{n-1} z^n$. By the above rule for computing radius of convergence, you can show $g$ has the same radius of convergence as $f$.
Next, pick z, h with |z| < r < R and |z + h| < r. Write $f(z) = S_n(z) + E_{n+1}(z)$, where $S_n$ is the partial sum and $E_{n+1}$ is the rest.
$$
\begin{eqnarray}
\dfrac{f(z+h) – f(z)}{h} – g(z) &=& \left(\dfrac{S_n(z+h) – S_n(z)}{h} – S_n'(z)\right) \\
&+& (S_n'(z) – g(z)) \\
&+& \dfrac{E_{n+1}(z+h) – E_{n+1}(z)}{h}
\end{eqnarray}
$$
The first term clearly converges to 0 as h -> 0 by definition of the derivative, as does the second term by definition of g as n -> infinity. Now,
$$
D_{n,h}(z) = \frac{E_{n+1}(z+h) – E_{n+1}(z)}{h} = \frac{1}{h}\displaystyle\sum_{k=n+1}^\infty a_k ((z+h)^k – z^k).
$$
Note that $(x^{m+1} – y^{m+1}) = (x – y)(x^m + x^{m-1}y + \dots + xy^{m-1} + y^m)$. So each term in the sum becomes $a_k h ((z+h)^{k-1} + \dots + z^{k-1})$. Since |z|, |z+h| < r, in magnitude this quantity is bounded by $k|h||a_k|r^{k-1}$. So $|D_{n,h}(z)| \leq \sum k |a_k| r^{k-1} \lt \infty$; as the tail of a convergent series (remember that g is absolutely convergent on the same disk), it also approaches zero as n approaches infinity.
Convergence of holomorphic functions
Theorem. Let $R$ be an open region, and suppose $f_n$ is a sequence of holomorphic functions on $R$ which converge uniformly to $f$. Then $f_n’$ converges uniformly to $f’$ on every compact subset of $R$.
Proof: By uniform convergence, we can find $M_n > 0$ such that $\sup_{z\in R} |f_n – f| < M_n$ and $M_n \to 0$.
Pick $\delta > 0$, and let $A_{\delta}$ be the set of all $z$ in $R$ which are at distance > $\delta$ to the boundary. We only need to show $f_n \to f$ uniformly on $A_{\delta}$.
For any $z \in A_{\delta}$, the disk $D$ of radius $\delta$ centered at $z$ lies in $R$. Then by Cauchy’s integral formula, we have
$$
\begin{eqnarray}
|f'(z) – f_n'(z)| &=& \left| \frac{1}{2\pi i} \int_{C_D} \frac{f(w) – f_n(w)}{(z – w)^2} dw \right| \\
&\leq& \frac{1}{2\pi} \frac{M_n * 2\pi\delta}{\delta^2} \\
&=& \frac{M_n}{\delta} \to 0
\end{eqnarray}
$$
This estimate depends only on $n$ and not on $z$, so the convergence is uniform.