I am faced with the following problem:
Suppose $f(z)$ and $g(z)$ have poles of order $m$ and $n$ respectively, at a point $z_{0} \in \mathbb{C}$ with $m \neq n.$ Show that $z_{0}$ is an isolated singular point of $f+g$ and find its type.
What I did was, first I said that if $f$ has a pole of order $m$ at $z_{0}$, then its Laurent expansion about $z_{0}$ is
$$ f(z) = \sum_{j=0}^{\infty}a_{j}(z-z_{0})^{j} + \frac{b_{1}}{z-z_{0}} + \frac{b_{2}}{(z-z_{0})^{2}} + \cdots + \frac{b_{m}}{(z-z_{0})^{m}}$$
Also, if $g$ has a pole of order $n$ at $z_{0}$, then its Laurent expansion about $z_{0}$ is
$$ g(z) = \sum_{j=0}^{\infty}c_{j}(z-z_{0})^{j} + \frac{d_{1}}{z-z_{0}} + \frac{d_{2}}{(z-z_{0})^{2}} + \cdots + \frac{d_{n}}{(z-z_{0})^{n}}$$
So, if $m < n$, then the Laurent expansion for $f+g$ is given by
$$ f(z) + g(z) = \sum_{j=0}^{\infty}(a_{j}+c_{j})(z-z_{0})^{j} + \frac{b_{1}+d_{1}}{z-z_{0}} +\cdots + \frac{b_{m}+d_{m}}{(z-z_{0})^{m}} + \frac{d_{m+1}}{(z-z_{0})^{m+1}} + \cdots + \frac{d_{n}}{(z-z_{0})^{n}}$$
which means that $f+g$ has a pole of order $n$ at $z_{0}$.
On the other hand, if $n < m$, then the Laurent expansion for $f+g$ is given by
$$ f(z) + g(z) = \sum_{j=0}^{\infty}(a_{j}+c_{j})(z-z_{0})^{j} + \frac{b_{1}+d_{1}}{z-z_{0}} +\cdots + \frac{b_{n}+d_{n}}{(z-z_{0})^{n}} + \frac{b_{n+1}}{(z-z_{0})^{n+1}} + \cdots + \frac{b_{m}}{(z-z_{0})^{m}}$$
which means that $f+g$ has a pole of order $m$ at $z_{0}$.
Is this all there is to this exercise?
Is $z_{0}$ also any other type of singular point?
Is what I did above sufficient to show that $f + g$ has an isolated singular point at $z_{0}$?
Let's say if I didn't know that $f$ had a pole or $g$ had a pole at $z_{0}$, but just an isolated singular point whose type was unspecified.
- Then, I know that $f$ is not defined analytically at $z_{0}$, but it is defined analytically in a deleted neighborhood of $z_{0}$, say $0 < |z-z_{0}| < \delta_{1}$.
- Also, I know that $g$ is not defined analytically at $z_{0}$, but is defined analytically in a deleted neighborhood of $z_{0}$, say $0 < |z-z_{0}| < \delta_{2}$.
- Is it true then to say that $f + g$ is defined analytically in a deleted neighborhood of $z_{0}$ whose radius $\delta = \min\{ \delta_{1}, \delta_{2} \}$?
- Why or why not?