I have been given the following function for which I am supposed to construct the laurent series around the singular point -1:
$$f(z)=\frac{z^2+4z+4}{z+1}$$
From the general formula I know that this implies:
$$f(z)=\sum_{n=-\infty}^\infty a_n*(z-(-1))$$
Since we are instructed to only use the geometric series I did the following subsitution:
$$x=z-1 \Rightarrow z=x+1$$
So the function takes the following form:
$$f=\frac{(x+1)^2+4(x+1)+4}{x+2}$$ which can be rewritten as:
$$f=\frac{(x^2+6x+10)}{2}*\frac{1}{1-(-\frac{x}{2})}$$
then applying the geometric series yields:
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}*x^{n+2}+6x^{n+1}+10x^n$$ from this expression one can do the backsubstitution.
Is this in any way the correct approach? I am unsure about the boundedness of the series I found, as far as I understand it, it should hold for $$0<|z|<2$$ and I don't know if I did the subsitution correctly in the first place.
Thanks for any helpful advice or correction!
The function \begin{align*} \color{blue}{f(z)=\frac{z^2+4z+4}{z+1}} \end{align*} has a simple pole at $z=-1$ and is analytic in $\mathbb{C}\setminus\{-1\}$. A Laurentexpansion at $z=-1$ can be obtained by expanding the numerator of $f$ in terms of $(z+1)$. We obtain \begin{align*} \color{blue}{f(z)}&=\frac{z^2+4z+4}{z+1}\\ &=\frac{(z+2)^2}{z+1}=\frac{(1+(z+1))^2}{z+1}\\ &=\frac{1+2(z+1)+(z+1)^2}{z+1}\\ &\,\,\color{blue}{=\frac{1}{z+1}+2+(z+1)} \end{align*}