TITLE: Questionable Proof in Visual Complex Analysis
QUESTION [5 upvotes]: I am currently reading the book "Visual Complex Analysis". It's a great book so far, but already in the beginning the proof of the identity theorem seems dubious. I mean, it's known from high-school algebra, that you're not allowed to divide by z if you set z = 0.
Isn't this proof completely wrong? Some friends I've asked even told me that this theorem is wrong in the reals, so wouldn't a correct proof have to use properties of the complex numbers? Is it even possible to prove this with elementary methods? I've seen proofs that use properties of holomorphic functions but haven't gotten that far in my book yet, so I have no experience with holomorphic functions.

REPLY [2 votes]: The argument here is, that you talk about a neighborhood (or even a set containing not only $0$):

If $$c_0 + c_1 z + c_2 z^2 + \dots = d_0 + d_1 z + d_2 z^2 + \dots$$
  for all z in a neighborhood of 0, then $c_j = d_j \ (\forall j \in \Bbb{N}_0)$

If these two only coincide in $z=0$ you can't argue $c_j = d_j$ (take for example $1 + z + z^2 + \dots = 1 + 2z + 2z^2 + \dots$ in $z=0$).
But since those two coincide in a whole neighborhood, you can simply apply cancellation laws. After the first step you are left with
$$z (c_1 + c_2 z + c_3 z^2 + \dots) = z(d_1 + d_2 z + d_3 z^2 + \dots)$$
For $z=0$ you can't conclude anything because you simply get the tautology $0=0$. But since they also coincide for values $z\neq 0$ basic cancellation leaves you with 
$$c_1 + c_2 z + c_3 z^2 + \dots = d_1 + d_2 z + d_3 z^2 + \dots$$
Now you plug $z=0$ in, get $c_1 = d_1$ and start again. This leaves you with $c_j = d_j$ for $z\neq 0$ but since power series are continuous they must also coincide for $z = 0$.
This proof also shows that your argument doesn't need the specific attributes of complex numbers and therefore uniqueness also holds for real valued power series.