Complex Analysis: Identity Theorem

[nateHI]

Homework Statement

Let f be a function with a power series representation on a disk, say [itex]D(0,1)[/itex]. In each case, use the given information to identify the function. Is it unique?
(a) [itex]f(1/n)=4[/itex] for n=1,2,[itex]\dots[/itex]
(b) [itex]f(i/n)=-\frac{1}{n^2}[/itex] for n=1,2,[itex]\dots[/itex]

A side question:
Is corollary 1 from my textbook given below just the Uniqueness Theorem?

Homework Equations

Corollary 1
Let [itex]g_1[/itex] and [itex]g_2[/itex] be holomorphic on the open connected set [itex]O[/itex], such that [itex]g_1=g_2[/itex] on a set that has an accumulation point within [itex]O[/itex]. Then [itex]g_1=g_2[/itex] throughout [itex]O[/itex].

The Attempt at a Solution

(a)
Consider [itex]g(z)=f(z)-4[/itex]. [itex]g[/itex] has a zero set [itex]\{1/n\}[/itex] that converges to [itex]0[/itex]. Therefore, by, the identity theorem
$$
g(z)=0\implies f(z)=4
$$
Finally, we note that [itex]f(z)=4[/itex] 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 [itex]f(z)[/itex] is unique.
(b)
Consider [itex]g(z)=f(z)-z^2[/itex]. [itex]g[/itex] has a zero set [itex]\{i/n\}[/itex] that converges to [itex]0[/itex]. Therefore, by the Identity Theorem,
$$
g(z)=0\implies f(z)=z^2
$$
Finally, we note that [itex]z^2[/itex] 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 [itex]f(z)[/itex] is unique.

 

[mfb]

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.

 

[nateHI]

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

 

[mfb]

Ah, okay.