Using Uniqueness Result for Analytic Functions 
I am reviewing for an Analysis qual and stumbled upon this question. In particular, I am having difficulties with part (ii). My attempt is the following:
Using the hint, let $\Omega = \mathbb{C}$, $S=\{1/n : n\in \mathbb{N}\}$, and $g(z)=z^2$. We have that since $S \subset \Omega$ and both $f$ and $g$ are entire, then $f$ and $g$ are analytic on $S$. Per the uniqueness result, if $g(z)=f(z)$ for all $z\in S$, we know that since $0$ is a limit point of $S$ that is in $\Omega$, then it must be the case that $g(z)=f(z)$ for all $z\in \Omega$. However, we are given that $|f(i)| =2$, yet $|g(i)| = 1$. So in this case, just because $|g(z)| = |f(z)|$ for all $z\in S$, we don't have $g(z)=f(z)$. My strategy is then to find different functions $g$ such that  $|g(z)| = |f(z)|$ for all $z\in S$ and $|g(i)|=2$. After finding all these different $g's$, I should have all the possible values of $|f(-i)|$ by just calculating $|g(-i)|$. However, I'm having trouble finding even a single function $g$ that satisfies these two conditions, much less finding all of them. Is there some systematic way I can go about finding these different $g$ functions?

First, if $f$ is any entire function, $\overline{f(\bar{z})}$ is always entire, because $\overline{\sum_{n=0}^\infty a_n \bar{z}^n} = \sum_{n=0}^\infty \bar{a_n}z^n,$ which converges exactly when $\sum_{n=0}^\infty a_n z^n$ does.
As $f$ is holomorphic at $0$ and not uniformly $0$, there is a unique integer $n \geq 0$ so $\lim_{z \to 0} \frac{f(z)}{z^n}$ is a nonzero complex number ($n$ is the order of the zero of $f$ at $0$). Our hypotheses show that $f$ has a zero of degree $2$ at $0$. So we can express $f(z) = z^2 h(z)$ for some entire function $h$, and our hypotheses show that $|h(z)| = 1$ for $z = 1/n, n \geq 1$. In particular, $|h(0)| = 1$. So there's a neighborhood of $0$ on which $h(z)$ is nonzero.
Now put $g(z) = \frac{1}{\overline{h(\bar{z})}}$, which holomorphic on a neighborhood of $0$, by part i. We observe that for $1/n$, $n\geq 1$, $$\frac{h(z)}{g(z)} = h(z) \overline{h(z)} = |h(z)|^2 = 1.$$ So by the identity principle, $g(z)$ and $h$ agree in a neighborhood of $0$, and we discover that in fact on all $z \in \mathbb{C}$, $h(z) = \frac{1}{\overline{h(\bar{z})}},$ or equivalently that $h(\bar{z}) = \frac{1}{\overline{h(z)}}$. For $|z| = 1$, $|f(z)| = |h(z)|$ and we see that for $|z| = 1$, $|f(\bar{z})| = \frac{1}{|f(z)|}$. So we conclude $|f(-i)| = 1/|f(i)| = 1/2.$
