I'm following Palka's Introduction to complex function theory, and in page 318 he states the theorem:
If neither of two functions $f$ and $g$ have worse than a pole at $z_0$ then $f+g$ doesn't have worse than a pole at $z_0$.
My question is, can we know the exact order of the pole of $f+g$ at $z_0$ by finding it's order at $f$ and $g$ separately?
This question came while studying the function $h(z)=f(z)+g(z)=\frac{z^2}{\exp(z^5)-1} - \frac{1}{z^3}$. Studying $f$ and $g$ separately I found that both have poles of order 3 at $z_0$, but I found it much harder to study the pole when $h(z)$ was written as $h(z)= \frac{z^5 -(\exp(z^5)-1)}{({\exp(z^5)-1})z^3}$.
The answer to your question is in the theorem below. But first a few lemmas.
Lemma 1. Let $a \in \mathbb{C}$, let $D \subseteq \mathbb{C}$ be an open set such that $a \in D$, and let $f : (D\setminus\{a\})\rightarrow\mathbb{C}$ be holomorphic. Then $f$ has a removable singularity at $a$ iff $\lim_{z\rightarrow a}(z-a)f(z) = 0$.
Proof. This is part of Riemann's Theorem. $\square$
Lemma 2. Let $a \in \mathbb{C}$, let $D \subseteq \mathbb{C}$ be an open set such that $a \in D$, let $f : (D\setminus\{a\})\rightarrow\mathbb{C}$ be holomorphic, and let $m \in \{1,2,3,\dots\}$. Then $f$ has a pole at $a$ of order $m$ iff the limit $\lim_{z\rightarrow a}(z-a)^mf(z) \in \mathbb{C}\setminus\{0\}$ (in particular, the limit exists).
Proof. Use a similar proof to the one given to Theorem 2.4 on p. 317 of An Introduction to Complex Function Theory by Bruce P. Palka (Springer, 1991). $\square$
Lemma 3. Let $a \in \mathbb{C}$, and let $D \subseteq \mathbb{C}$ be an open set such that $a \in D$. For every $m \in \{1,2,3,\dots\}$ the following statement holds: every holomorphic $f : (D\setminus\{a\})\rightarrow\mathbb{C}$ satisfying that $\lim_{z\rightarrow a}(z-a)^mf(z) = 0$ (in particular, the limit exists), either has a removable singularity at $a$, or, if $m > 1$, has a pole of order $< m$ at $a$.
Proof. We proceed by induction on $m$.
Base case: $m = 1$. The conclusion follows from lemma 1.
Inductive step. Assume the claim holds for some $m \in \{1,2,3,\dots\}$. Let $f : (D\setminus\{a\})\rightarrow\mathbb{C}$ be holomorphic and suppose it satisfies that $\lim_{z\rightarrow a}(z-a)^{m+1}f(z) = 0$ (in particular, the limit exists).
Define the function $g:(D\setminus\{a\})\rightarrow\mathbb{C}$ as follows: $g(z) = (z-a)^mf(z)$.
Then $\lim_{z\rightarrow a}(z-a)g(z) = \lim_{z\rightarrow a}(z-a)^{m+1}f(z) = 0$. Therefore, since $g$ is holomorphic (being the product of two holomorphic functions), lemma 1 implies that $g$ has a removable singularity at $a$. Hence, in particular, the limit $\lim_{z\rightarrow a}(z-a)^mf(z) = \lim_{z\rightarrow a}g(z)$ exists and belongs to $\mathbb{C}$. Denote this limit by $L$.
If $L = 0$, the desired conclusion follows from the inductive hypothesis; otherwise, the desired conclusion follows from lemma 2. $\square$
Theorem. Let $a \in \mathbb{C}$, let $D \subseteq \mathbb{C}$ be an open set such that $a \in D$, and let $f, g : (D\setminus\{a\})\rightarrow\mathbb{C}$ be holomorphic functions such that $f$ has a pole of order $m$ at $a$, and $g$ a pole of order $n$ there, with $m, n \in \{1,2,3,\dots\}$. Then
Proof.
Define $F = \lim_{z\rightarrow a}(z-a)^mf(z)$, and $G = \lim_{z\rightarrow a}(z-a)^nf(z)$. By lemma 2, $F$ and $G$ are well-defined and belong to $\mathbb{C}\setminus\{0\}$.
Assume w.l.g. that $m < n$. Then $$ \begin{multline*} \lim_{z\rightarrow a}(z-a)^n\big(f(z)+g(z)\big)\\ = \lim_{z\rightarrow a}(z-a)^{n-m}\lim_{z\rightarrow a}(z-a)^mf(z) + \lim_{z\rightarrow a}(z-a)^ng(z)\\ = 0F+G \in \mathbb{C}\setminus\{0\}. \end{multline*} $$
By lemma 2, $f+g$ has a pole of order $n = \max(m,n)$ at $a$.
We have $$ \begin{multline*} \lim_{z\rightarrow a}(z-a)^n\big(f(z)+g(z)\big) \\= \lim_{z\rightarrow a}(z-a)^mf(z) + \lim_{z\rightarrow a}(z-a)^ng(z)\\ = F + G. \end{multline*} $$
If $F + G = 0$, the desired conclusion follows from lemma 3; otherwise, from lemma 2. $\square$