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*}

  • $\begingroup$ At last I understand :) $\endgroup$ – paranoidhominid Jul 6 '16 at 13:47
  • 1
    $\begingroup$ @paranoidhominid: Good to see the answer is useful! :-) $\endgroup$ – Markus Scheuer Jul 6 '16 at 13:51

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