# Derive a Laurent series for the function $2z/(z+j)$

First of all, I apologize for the none mathematical notations. I've only ever hanged around Stackoverflow, and never learnt how to type Mathmatical notations. It would be great if someone could teach me. The good news is my question isn't too long.

I'm a beginner learning Taylor and Laurent series. I know how to find the Laurent series for simple functions such as $1/(z+5)$ by rearranging it into similar form as the geometric series $1/(1-z)$.

However, I came across this question,

Derive a Laurent series for $f(z)=\dfrac{2z}{z+j}$ about the centre $z=-j$.

I tried rearranging it to $f(z)=1/((1/z)(1+(j/z)))$ but it doesn't seem to give me the right answer.

Maybe I'm not understanding something, or maybe I'm just making a silly mathematic error. But any help would be much appreciated.

Recall that the Laurent series for $f(z)$ centered around $-j$ is written as $$f(z) = \cdots + \dfrac{a_{-2}}{(z+j)^2} + \dfrac{a_{-1}}{(z+j)} + a_0 + a_1(z+j) + a_2 (z+j)^2 + \cdots$$ In our case, we are given $f(z) = \dfrac{2z}{z+j}$. \begin{align} f(z) &= \dfrac{2z}{z+j} = \dfrac{2z+2j - 2j}{z+j} & \text{(Adding and subtracting $2j$ in the numerator)}\\ & = \dfrac{2z+2j}{z+j} - \dfrac{2j}{z+j}\\ & = 2 \dfrac{z+j}{z+j} - \dfrac{2j}{z+j}\\ & = 2 - \dfrac{2j}{z+j} \end{align} The above is the Laurent series centered around $j$ with $$a_n = \begin{cases} -2j & n=-1\\ 2 & n=0\\ 0 & n \in \mathbb{Z} \backslash\{0,-1\} \end{cases}$$