# Understanding proof regarding uniqueness of power series representation

Theorem 3.2 in Lang's Complex Analysis has part (a) and part (b), it's actually part (b) that shows the uniqueness of power series, but I'm having trouble completely understanding part (a). Here is the theorem of part (a), the proof, and my question about it:

$$Theorem:$$ Let $$f(T)= \Sigma a_nT^n$$ be a non-constant power series with a non-zero radius of convergence where $$a_n \in \mathbb{C}$$. If $$f(0)=0$$, then $$\exists s>0$$ such that $$f(z) \neq 0$$ for all $$z$$ with $$|z| \leq s$$, and $$z \neq 0$$

$$Proof:$$ We can write:

$$f(z)=a_mz^m$$ + higher terms (with $$a_m \neq 0$$).

$$=a^mz^m(1+b_1z+b_2z+b_3z+...)$$

$$=a_mz^m(1+h(z))$$

Where $$h(z)$$ is a power series having a non-zero radius of convergence, and zero constant term. For all sufficiently small $$|z|$$, the value $$|h(z)|$$ is small and hence $$1+h(z) \neq 0$$. If $$z \neq 0$$, then $$a_mz^m \neq 0$$. This proves (a).

Okay, so actually I understand this proof and theorem, but what I'm confused about is why we need to include the "if $$f(0)=0$$, then ..."

It seems like the same proof would work even if $$f(0) \neq 0$$:

$$f(z) = a_0+a_1z+a_2z^2+....$$

$$=a_0(1+b_1z+b_2z^2+....)$$

$$=a_0(1+h(z))$$

Where for small enough $$z$$ we have $$1+h(z) \neq 0$$ and $$a_0 \neq 0$$ by assumption. But this can't be correct, else it would seem that power series would never have zeros....

And maybe I'm just mixing up my logic a little bit, and that when they say "If $$f(0) = 0$$, then.." means more like, hey look, we can do this even when $$f(0)=0$$" more than thinking about the $$f(0)=0$$ to be some sort of necessity condition.

Anyway, I think the main point of this theorem is that zero's of analytic functions are isolated, but I am still a bit confused. I hope this is a bit clear. Any insight is greatly appreciated!!

• Part (b) requires exactly that to be used and then end at once the argument of the theorem...! Commented Apr 20, 2020 at 14:57

You're right that the same conclusion would be true if $$f(0)\ne 0$$, but it's not really the same proof, it's just continuity of $$f$$ at $$0$$.
It does not mean power series do not have zeros. Consider the simple example $$f(z) = z$$. For every complex number $$a$$ there is $$s > 0$$ such that $$f$$ has no zeros in $$\{z: 0 < |z-a| \le s\}$$. For $$a = 0$$ every $$s > 0$$ will work. For $$a \ne 0$$ you just need $$s < |a|$$. The point is that since $$s$$ depends on $$a$$, you can still have the zero at $$z=0$$, which is outside of all the punctured disks $$\{z: 0 < |z-a| \le s\}$$ for $$a \ne 0$$.