How to prove the partial derivative of a holomorphic function is holomorphic and obtained by differentiating the series term by term? Let $U\subset\mathbb{C}^n$ be open. Suppose $f:U\rightarrow\mathbb{C}$ is holomorphic, then by the definition, $f$ can be expressed by an absolutely and uniformly convergent power series,
$$f(\mathbf{z})=\sum_{I\in\mathbb{Z}^n_{\geq 0}}\alpha(I)(\mathbf{z}-\mathbf{z_0})^I=\sum_{I\in\mathbb{Z}^n_{\geq 0}}\alpha(I)(z_1-z_{01})^{i_1}\cdots(z_n-z_{0n})^{i_n},$$
in some neighbourhood of $\mathbf{z_0}$.
Is it true that $\frac{\partial f}{\partial z_i}(\mathbf{z})$ is holomorphic? Can the partial derivative of $f$ be obtained by differentiating the series term by term, which is
$$\frac{\partial f}{\partial z_i}(\mathbf{z})=\sum_{I\in\mathbb{Z}^n_{\geq 0}}\frac{\partial}{\partial z_i}\alpha(I)(\mathbf{z}-\mathbf{z_0})^I?$$
Thanks for your answer.
 A: Sketch of proof:
The key idea is that uniform limits do not, in general, commute with derivatives but they do commute with integrals.
Take any holomorphic function.  Repeatedly apply Cauchy's Integral Formula to give an integral representation for the function on a polydisk.  Check that the conditions for differentiation under the integral sign apply to obtain an integral representation of the derivative of the function.
Once you have an integral representation of holomorphic functions, you can interchange uniform limits with integrals to obtain the result you are after.
Now I should mention that the result on term by term differentiation of power series is true even in the real case and can be proven without recourse to integral representation theorems.  The arguments are kind of painful though.  I think my outline above is the least painful way to get the result you want, assuming you are confident in your single variable complex analysis.
A: Lets assume $i = 1$, Since $f(z)$ is absolutely convergent, for a fixed $z_2,...,z_n$, we have that $f(z_1) = \sum_i a_i(z_2,...,z_n) \times (z_1-z_{01})^i$. Now use the proof given for one-variable differentiation (http://www.math.ualberta.ca/~isaac/math309/ss17/diff.pdf) and take the derivative inside and get $\frac{\partial f(z_1)}{\partial z_1} = \sum_i a_i(z_2,...,z_n) \times \frac{\partial (z_1-z_{01})^i}{\partial z_1} = \sum_i a_i(z_2,...,z_n) \times i(z_1-z_{01})^{i-1}$.
So the assumptions in the above are whether we can re-arrange the terms in the series. This can be done since we have absolute convergence (See: A rearrangement of an absolutely convergent complex series is also absolutely convergent).
We further assume there is a neighbourhood of $z_{01}$ for $z_1$ in which the series absolutely converges for every fixed $z_2,...,z_n$. This is also true.
