Questionable Proof in Visual Complex Analysis

The theorem is true, and the proof is valid. However, it does omit a little step which is subtle. We have, thanks to the first part of the proof that $$\sum_{k=1}^\infty c_k z^k=\sum_{k=1}^\infty d_k z^k\\ z\left(\sum_{k=1}^\infty c_k z^{k-1}\right)=z\left(\sum_{k=1}^\infty d_k z^{k-1}\right)$$

Here comes the subtle part: for every $z\neq 0$, we have that

$$\sum_{k=1}^\infty c_k z^{k-1}=\sum_{k=1}^\infty d_k z^{k-1}$$

Since the two power series are continuous and are equal on a punctured neighbourhood of $0$, they must be equal at $0$ too. In fact:

$$c_1=\lim_{z\to 0}\sum_{k=1}^\infty c_k z^{k-1}=\lim_{z\to 0}\sum_{k=1}^\infty d_k z^{k-1}=d_1$$

The rest of the proof follows similarly

As a side note, the identity principle, is a feature of analytic functions in general and does not depends on their domain being $\mathbb{R}$ or $\mathbb{C}$

As I mentioned, this proof lacks of rigour. This is an intentional choice of the author through the book, as he states in the introduction:

"My book will no doubt be flawed in many ways of which I am not yet aware, but there is one "sin" that I have intentionally committed, and for which I shall not repent: many of the arguments are not rigorous, at least as they stand."

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.