I need to find both a Laurent and a Taylor expansion.
$$f(z)=\frac{z}{(z-1)(z-2)} = \frac{-1}{(z-1)}+\frac{2}{(z-2)}$$
If I choose $z_0=0$
$$f(z)=\frac{1}{(1 + z)} – \frac{4}{\left(1 – \frac{z}{4}\right)}$$
$$f(z)=\sum_{n}^{\infty}(-1)^n{z^n} – 4\sum_{n}^{\infty}(\frac{z}{4})^n$$
Which is a Taylor series.
What value of $z_0$ would you pick for a Laurent series?
Best Answer
It is sufficient to consider a Laurent expansion around $z=0$. Depending on the region which you then choose you obtain a principal part of a Laurent expansion respectively an expansion as Taylor series.
A Tayor series expansion of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\ &=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n \end{align*} The principal part of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n} =-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n} \end{align*}