# Questions tagged [laurent-series]

This tag is for questions about finding a Laurent series of functions and their convergence. The Laurent series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

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### Laurent Series of $\frac{1}{z(1-z)}$ in neighborhood of $z=1$ and $z=0$

So, the question is: Laurent Series of $\frac{1}{z(1-z)}$ in neighborhood of $z=1$ and $z=0$. I know I can find Laurent series' all over MSE, but in an effort to build my own intuition, and to see the ...
25 views

### Determining Laurent series on a region [closed]

$g(z)= \frac { \sinh2z } { (2-z)^2 }$ , $H = \{ z \in \mathbb C : 0 < |z-2| < \infty \}$. Show that $g(z)$ function has Laurent series expansion on $H$ region and determine that Laurent series....
76 views

### Show that $\sum_{k=0}^{\infty} \frac{k^2+3k+2}{2} z^k = \frac{1}{(1-z)^3}$, without using differentiation

Show that, $$\sum_{k=0}^{\infty} \frac{k^2+3k+2}{2} z^k = \frac{1}{(1-z)^3}$$ where $z \in \mathbb{C}, |z|< 1$ Well, I have figured out that is a Laurent series I have watched 3 videos in the ...
18 views

### Does a multivariate series expansion exist for this function?

I have the following function that I was hoping to simplify in some way $$\displaystyle f(x,y)=\frac{1+x}{1+y+\sqrt{(1-y)^2+y(1-\frac{1}{x})^2}}$$ defined for $0 < {x,y} \leq 1$ and was wondering ...
32 views

### Finding the Laurent Series of $f(z)=\frac{1}{(z^2+1)^2}$

I'm trying to determine the residue at $i$ of $f(z)=\frac{1}{(z^2+1)^2}$. My first attempt was to transform this into a Laurent Series: $$f(z)=\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2}$$ ...
41 views

### Why is $\operatorname{res}(0, \cos(\frac{1}{z}))=0$?

The residue of a function $f$ represented by a Laurent Series in complex analysis is defined as the coefficient at $n=-1$ of the series ($a_{-1}$). Giving the definition of the cossine in complex ...
Given the equation, $$\sum _{i=0}^{n}a_{i}x^{i}=\sum _{j=1}^{m}b_{j}x^{-j}$$ solve for x. I'd typically solve this by multiplying across by $x^{m}$, then shifting everything to one side and finding ...