I have the following question, and I am struggling to interpret what it wants me to show.
Let $U$ be an open subset of the complex plane, let $a \in U$ and let $U'= U \backslash \{a\}$. Assume that $f$ is an analytic function on $U'$ with $|f(z)| \to \infty$ as $z \to a$. By considering the Laurent series of $g(z) = 1/f(z)$ at $a$, classify the singularity of $f$ at a in terms of the Laurent coefficients. [You may assume that a continuous function on $U$ that is analytic on $U'$ is analytic on $U$.]
I'm really confused as to what this wants me to show. I know that the singularity at $a$ should be a pole, but isn't a pole just defined by the fact that $\lim_{z \to a} |f(z)| = \infty$?
To show it is a pole, though, I could try to show that the Laurent series of $f$ at $a$
$$\sum_{n= -\infty}^{\infty} c_n(z-a)^n$$
has $c_n=0$ for $n$ sufficiently large.
Considering $g(z)=\frac{1}{f(z)}$, we'd want $\lim_{z \to a} g(z) = 0$, so the Laurent series is
$$\sum_{n= \color{red}{0}}^{\infty} c'_n(z-a)^n$$
I'm then not sure how to relate the Laurent series of $g$ to that of $f$. An earlier part of the question required me to state the formula
$$c_n=\frac{1}{2 \pi i} \int_{\partial D_{\rho}(a)} \frac{f(z)}{(z-a)^{n+1}} dz,$$
for some $\rho$, where $D_{\rho}(a)$ is the disk of radius $\rho$ centred at $a$. I attempted to write this as
$$c_n=\frac{1}{2 \pi i} \int_{\partial D_{\rho}(a)} \frac{1}{g(z)(z-a)^{n+1}} dz$$
and plug in the Laurent series for $g$ into this, to no avail.
As I said, I'm not even sure if I am interpreting this question correctly since I am confused on what exactly it is I am being asked to do here. What should be the correct approach? I'm also not sure of why it emphasises $U$ and $U'$, and how I should incorporate that into this.
It's not you. There is just a disconnect between the question and the notes. It's possible that the question was written by a different instructor (perhaps for an earlier year) who didn't follow the notes.
If we are sticking to your notes, then by definition (p.38), $f$ has a pole at $a$. That's it.
Judging from the wording, I guess the person who wrote the question used the following equivalent definition in terms of Laurent series:
Definition: Consider $f$ with Laurent series expansion $f(z) = \sum_{k = -\infty}^\infty c_k (z - a)^k$ (by theorem, p.40, always exists). (i) If $c_{-1} = c_{-2} = \dots = 0$, then $f$ has a removable singularity at $a$; (ii) If $c_{-N} \neq 0$ for some $N > 0$, and $c_{-N - 1} = c_{-N - 2} = \dots = 0$, then $f$ has pole of order $N$ at $a$; (iii) If for any $k_0$, there is $k < k_0$ with $a_k \neq 0$, then $f$ has an essential singularity at $a$.
If we took this as the definition of removable/pole/essential singularity, then the question would make some sense. Now we can interpret the question as follows: Given that $|f(z)| \rightarrow \infty$ as $z \rightarrow a$, show that the Laurent series $f(z) = \sum_{k = -\infty}^\infty c_k (z - a)^k$ satisfies condition (ii) in our definition above.
The hard part was figuring out what the question meant. Solving it is relatively simple, as you already figured out the key step:
Take $g(z) = 1/f(z)$. Since $|f(z)| \rightarrow \infty$ as $z \rightarrow a$, we have $g(z) \rightarrow 0$ as $z \rightarrow a$. By that last remark in your question [You may assume that a continuous function on $U$ that is analytic on $U′$ is analytic on $U$.] we conclude that $g(z)$ is analytic on $U$, with some power series $\sum_{k = 0}^\infty b_k (z - a)^k$.
Since $g(a) = 0$, we can pull out a factor of $(z - a)^N$, where $N$ is the order of the zero. So $g(z) = (z - a)^N h(z)$, where $h(0) \neq 0$.
Now $f(z) = \frac{1}{(z - a)^N} \frac{1}{h(z)}$. Since $h(0) \neq 0$, $\frac{1}{h(z)}$ is analytic around $z = a$, with power series representation $\frac{1}{h(z)} = \sum_{k = 0}^\infty d_k (z - a)^k$, where $d_0 \neq 0$. Finally, the Laurent series of $f$ is $f(z) = \frac{1}{(z - a)^N} \sum_{k = 0}^\infty d_k (z - a)^k = \frac{d_0}{(z - a)^N} + \frac{d_1}{(z - a)^{N - 1}} + \dotsb$. By our definition above, $f$ has a pole of order $N$ at $a$.