What theorem tells us that a function has a power series expansion around a point? I'm trying to show that $(1+z)^a$ has power series expansion around $z=0$. What theorem could I use to directly prove this?
 A: A holomorphic function (defined and $\mathbb C$-differentiable on an open subset of $\mathbb C$) necessarily has a power series expansion around every point of its domain. The radius of convergence is at least the distance to the exterior of its domain.
That's the case of your function, if $a$ is a positive integer, that's a polynomial which is obviously holomorphic on $\mathbb C$. If it's a negative integer, holomorphic on $\mathbb C^*$, and if it's an arbitrary real, then you need to define it using a complex logarithm, making it holomorphic on $\mathbb C\setminus \mathbb R_-$.
A: That a function which is holomorphic in a domain $\Omega$ has a power series expansion in every disk $B_R(a) \subset \Omega$ is an immediate consequence of Cauchy's integral formula.
This can be seen as follows: For $0 < r < R$ and $|z-a| < r$ is
$$
 f(z) = \frac{1}{2 \pi i}\int_{|w-a|=r} \frac{f(w)}{w-z} \, dw \, .
$$
Now
$$
\frac{1}{w-z} =  \sum_{n=0}^\infty \frac{(z-a)^n}{(w-a)^{n+1}}
$$
converges uniformly on $|w-a|=r$, so that integration and summation can be exchanged:
$$ \tag{$*$}
 f(z) = \sum_{n=0}^\infty a_n (z-a)^n 
$$
where
$$
 a_n =  \frac{1}{2 \pi i }\int_{|w-a|=r} \frac{f(w)}{(w-a)^{n+1}} \, .
$$
It follows that $f$ can be expanded into a power series in $B_r(a)$.
This holds for all $0 < r < R$ and power series coefficients are uniquely determined. Therefore $(*)$ holds in the disk $B_R(a)$.
