# the laurent series expansion [closed]

Please, I need someone to help me to solve this problem, thanks. Let $$f(z) = \frac{2z+1}{z(z+i)},$$

1. Find the laurent series expansion for $f(z)$ in $ann(0;1,∞).$
2. Find the laurent series expansion for $f(z)$ in $ann(i;0,1).$

## closed as unclear what you're asking by zz20s, user91500, Watson, C. Dubussy, choco_addictedApr 25 '16 at 8:27

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## 1 Answer

The function

\begin{align*} f(z)&= \frac{2z+1}{z(z+i)}=\frac{2+i}{z+i}-\frac{i}{z} \end{align*} has two simple poles at $0$ and $-i$.

First part: Laurent expansion in annulus with center $z=0$, inner radius $1$ and outer radius $\infty$.

Since we want to find a Laurent expansion at $z=0$, we look at the poles $0$ and $-i$ and see they determine two regions.

\begin{align*} 0<|z|<1,\qquad\quad 1<|z|,\qquad\quad \end{align*}

• The first region $0<|z|<1$ is a punctured disc with center $0$, radius $1$ and the pole $-i$ at the boundary of the disc. Here we have a representation of the fraction with pole $0$ as principal part of a Laurent series at $z=0$, while the fraction with pole $-i$ admits a representation as power series.

• The second region $|z|>1$ containing all points outside the disc with center $0$ and radius $1$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power 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}}\\ &=\frac{1}{a}\sum_{n=0}^{\infty}\left(-\frac{1}{a}\right)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}\left(-\frac{a}{z}\right)^n =\sum_{n=0}^{\infty}(-a)^n\frac{1}{z^{n+1}}\\ &=\sum_{n=1}^{\infty}(-a)^{n-1}\frac{1}{z^{n}}\tag{1} \end{align*}

In this example we are interested in the second region. So, we take the principal part (1) and get \begin{align*} f(z)&=\frac{2+i}{z+i}-\frac{i}{z}\\ &=(2+i)\sum_{n=1}^{\infty}(-i)^{n-1}\frac{1}{z^{n}}-\frac{i}{z}\\ &=\frac{2}{z}+(2+i)\sum_{n=2}^{\infty}(-i)^{n-1}\frac{1}{z^{n}}\qquad\qquad z\in ann(0;1,\infty)\\ \end{align*}

Second part: Laurent expansion with center $z=i$ in disc with radius $1$.

Since we want to find a Laurent expansion at $z=i$, we look at the poles $0$ and $-i$ and see they determine three regions.

\begin{align*} 0<|z-i|<1,\qquad\quad 1<|z-i|<2,\qquad\quad 2<|z-i|,\qquad\quad \end{align*}

• The first region $0<|z-i|<1$ is a disc with center $i$,radius $1$ and the pole $0$ at the boundary of the disc. Here we have a representation of the fraction with poles $0$ and $-i$ as power series.

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

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

A power series expansion of $\frac{1}{z+a}$ at $z=i$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{(a+i)+(z-i)}=\frac{1}{a+i}\cdot\frac{1}{1+\frac{z-i}{a+i}}\\ &=\frac{1}{a+i}\sum_{n=0}^{\infty}\left(-\frac{z-i}{a+i}\right)^n\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{(a+i)^{n+1}}(z-i)^n\tag{2}\\ \end{align*} The principal part of $\frac{1}{z+a}$ at $z=i$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{(z-i)+(a+i)}=\frac{1}{z-i}\cdot\frac{1}{1+\frac{a+i}{z-i}}\\ &=\frac{1}{z-i}\sum_{n=0}^{\infty}(-1)^n\left(\frac{a+i}{z-i}\right)^n\\ &=\sum_{n=0}^{\infty}(-1)^n(a+i)^n\frac{1}{(z-i)^{n+1}}\\ &=\sum_{n=1}^{\infty}(-1)^{n-1}(a+i)^{n-1}\frac{1}{(z-i)^{n}} \end{align*}

In this example we are interested in the first region. So, we take the power series part (2) and get \begin{align*} f(z)&=\frac{2+i}{z+i}-\frac{i}{z}\\ &=(2+i)\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2i)^{n+1}}(z-i)^{n}-i\sum_{n=0}^{\infty}\frac{(-1)^n}{i^{n+1}}(z-i)^n\\ &=\frac{2+i}{2i}\sum_{n=0}^{\infty}\frac{i^n}{2^n}(z-i)^{n}-\sum_{n=0}^{\infty}i^n(z-i)^n\\ &=\sum_{n=0}^{\infty}\left(\frac{1-2i}{2^{n+1}}-1\right)i^n(z-i)^n\qquad\qquad z\in ann(i;0,1)\\ \end{align*}