0
$\begingroup$

Let $D$ be an open subset of $\mathbb{C}$ and $f,g : D \rightarrow \mathbb{C}$ be meromorphic, both having a pole at $z_0$. If I know the Laurent series of $f$ and $g$ around $z_0$, how to calculate the Laurent series of $\frac{f}{g}$ around $z_0$? I made up an example and couldn't transform it into the standard form to see the coefficients. This is my example with $z_0=0$: $$f(z)=z^{-2}+3z+2z^2$$ $$g(z)=z^{-2}+4z+8z^3$$

Further, if $f$ and $g$ have the same principal part (like in my example), then I want to show that $\frac{f}{g}$ doesn't have a principal part. How can I show that?

$\endgroup$
1
$\begingroup$

Your second question sounds like a use of Morera's Theorem: If a function is analytic on $D$ except a single point $z_0$ where it is continuous, then it is analytic at that point as well.

So, $f/g$ won't have a principal part at $z_0$ because (with the assumption that they have the same pole of the same order) then the following limit will exist, $$\lim_{z\to z_0} \frac{f}{g} .$$ Like in your example, $$\lim_{z\to 0} \frac{f}{g} =1.$$

Your fist question, about how to find the Lauren series of a quotient, is much more difficult to answer. Typically, special cases are known and techniques are learned, but a general form might prove quite tedious. If you aren't allowing essential singularities, a fairly simple technique can be developed, let $$f(z)=\sum_{k=-n}^\infty a_k (z-z_0)^k\ \ \ \text{and}\ \ \ g(z)=\sum_{k=-m}^\infty b_k (z-z_0)^k,$$ where $a_{-n}$ and $b_{-m}$ are not zero. Then $$(z-z_0)^nf(z)\ \ \ \text{and}\ \ \ (z-z_0)^mg(z)$$ are analytic functions in a neighborhood of $z_0$, that are not zero at $z_0$. Hence, their quotient is an analytic function at $z_0$ and thus a taylor series can be found, like, $$\frac{f(z)}{g(z)}=(z-z_0)^{m-n} \left(\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right) = (z-z_0)^{m-n} \sum_{k=0}^\infty \left(\left.\frac{d^k}{dz^k}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}\right)(z-z_0)^k.$$ Hence, $$\frac{f(z)}{g(z)} = \sum_{k=0}^\infty \left(\left.\frac{d^k}{dz^k}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}\right)(z-z_0)^{m-n+k}. $$ So, if you were asked in integrate this, you would simply need to find the $k$ where $m-n+k=-1$ and then use the coefficient at that part and calculate, $$\int_C \frac{f(z)}{g(z)}dz = 2\pi i \cdot \left.\frac{d^{n-m-1}}{dz^{n-m-1}}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}.$$ If you actually calculate this out for specific $n$ and $m$ then you obtain many residue formulas found in textbooks. Take a look at Marsden - Complex Analysis - Page 250, which presents a table dedicated to this very concept.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.