Complex Analysis: Identity Theorem

1. Mar 4, 2015

nateHI

1. The problem statement, all variables and given/known data
Let f be a function with a power series representation on a disk, say $D(0,1)$. In each case, use the given information to identify the function. Is it unique?
(a) $f(1/n)=4$ for n=1,2,$\dots$
(b) $f(i/n)=-\frac{1}{n^2}$ for n=1,2,$\dots$

A side question:
Is corollary 1 from my textbook given below just the Uniqueness Theorem?
2. Relevant equations
Corollary 1
Let $g_1$ and $g_2$ be holomorphic on the open connected set $O$, such that $g_1=g_2$ on a set that has an accumulation point within $O$. Then $g_1=g_2$ throughout $O$.

3. The attempt at a solution
(a)
Consider $g(z)=f(z)-4$. $g$ has a zero set $\{1/n\}$ that converges to $0$. Therefore, by, the identity theorem
$$g(z)=0\implies f(z)=4$$
Finally, we note that $f(z)=4$ satisfies the conditions given in the problem statement for part (a). Also, in solving the first part of this problem we've demonstrated the hypothesis for Corollary 1, thereby showing that $f(z)$ is unique.
(b)
Consider $g(z)=f(z)-z^2$. $g$ has a zero set $\{i/n\}$ that converges to $0$. Therefore, by the Identity Theorem,
$$g(z)=0\implies f(z)=z^2$$
Finally, we note that $z^2$ satisfies the conditions given in the problem statement for part (b). Also, in solving the first part of this problem we've demonstrated the hypothesis for Corollary 1, thereby showing that $f(z)$ is unique.

2. Mar 4, 2015

Staff: Mentor

I'm not sure what you mean with "zero set [...] that converges to 0" - a set does not converge, and what is a zero set?
You can probably skip some steps you make twice, but the proof is fine.

3. Mar 4, 2015

nateHI

A zero sequence would be a more apt descriptor then. So, a zero sequence is a sequence $\{z_k\}$ such that $f(z_k)=0$ for all k and where $\lim_{k\to\infty} z_k\to a$. In my case a=0.

4. Mar 4, 2015

Ah, okay.