Finding Laurent and Taylor series

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?

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.

The function

\begin{align*} f(z)&= \frac{-1}{z-1}+\frac{2}{z-2}\\ \end{align*} has two simple poles at $$1$$ and $$2$$.

We look at the poles $$1$$ and $$2$$ and see they determine three regions.

\begin{align*} |z|<1,\qquad\quad 1<|z|<2,\qquad\quad 2<|z| \end{align*}

• The first region $$|z|<1$$ is a disc with center $$0$$, radius $$1$$ and the pole $$1$$ at the boundary of the disc. In the interior of this disc all two fractions with poles $$1$$ and $$2$$ admit a representation as Taylor series at $$z=0$$.

• The second region $$1<|z|<2$$ is the annulus with center $$0$$, inner radius $$1$$ and outer radius $$2$$. Here we have a representation of the fraction with pole $$1$$ as principal part of a Laurent series at $$z=0$$, while the fraction with pole at $$2$$ admits a representation as power series.

• The third region $$|z|>2$$ containing all points outside the disc with center $$0$$ and radius $$2$$ admits for all fractions a representation as principal part of a Laurent series at $$z=0$$.

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*}

• At last I understand :) – paranoidhominid Jul 6 '16 at 13:47
• @paranoidhominid: Good to see the answer is useful! :-) – Markus Scheuer Jul 6 '16 at 13:51