Find the Taylor series about $z=0$ and the Laurent series about $z=-3$

Question

• Let $f(z)=\frac{10z}{z^2+z-6}$, find the coefficient of $z$ in the Taylor series of $f$ expanded about $z=0$ and state the open set in $\mathbb C$ where the series converges.

• Find the Laurent series of $f$ about $z=-3$ and state the open set where this series converges.

• For the first part:

$$\begin{align}f(z)&=\frac{6}{z+3}+\frac{4}{z-1}\\ &=\frac{6}{3}\frac{1}{1+\frac{z}{3}}-4\frac{1}{2-z}\\ &=2\frac{1}{1-(-\frac{z}{3})}-2\frac{1}{1-\frac{z}{2}}\end{align}$$

So Taylor series would be

$$2\sum_{n=0}^{\infty}\left(-\frac{z}3\right)^n-2\sum_{n=0}^{\infty}\left(\frac{z}2\right)^n$$

Where $|-\frac{z}{3}|<1$ and $|\frac{z}{2}|<1$, so $|z|<3$ and $|z|<2$, is this correct so far and how do I deduce the open set from this? Would it be $B_3(0)\setminus B_2(0)$?

• For the second part:

Let $z=-3+w$ and input into $f$,

$$\begin{align}f(z)&=\frac{6}{(-3+w)+3}+\frac{4}{(-3+w)-2} \\ &=\frac{6}{w}+\frac{4}{-5+w}\\ &=\frac{6}{1-1+w}+\frac{-4}{5-w}\end{align}$$

....and then I get a little lost from there.

Any corrections, tips or solutions would be great!!

Answer 1

We consider the function \begin{align*} f(z)&=\frac{10z}{z^2+z-6}\\ &= \frac{6}{z+3} +\frac{4}{z-2}\\ \end{align*}

First part: The function $f$ is to expand around the center $z=0$ as Taylor series.

Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish three regions of convergence \begin{align*} D_1:&\quad 0\leq |z|<2\\ D_2:&\quad 2<|z|<3\\ D_3:&\quad |z|>3 \end{align*}

• The first region $D_1$ is a disc with center $z=0$, radius $2$ and the pole at $z=2$ at the boundary of the disc. It admits for both fractions a representation as power series.

• The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole at $z=2$ a representation as principal part of a Laurent series and for the fraction with pole at $z=3$ a power series.

• The region $D_3$ contains all points outside the disc with center $z=0$ and radius $3$. It admits for both fractions a representation as principal part of a Laurent series.

We are interested in an expansion as Taylor series and observe the region $D_1$ is the corresponding open set of convergence with the representation of $f$ as Taylor series as stated by OP:

\begin{align*} f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\ &= 2\sum_{n=0}^\infty\left(-\frac{z}{3}\right)^n-2\sum_{n=0}^\infty\left(-\frac{z}{2}\right)^n \end{align*}

Second part: The function $f$ is to expand around the center $z=-3$ as Laurent series and the region of convergence is to determine.

Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish two regions of convergence when expanding around $z=-3$: \begin{align*} D_4:&\quad 0< |z+3|<5\\ D_5:&\quad |z+3|>5 \end{align*}

• The region $D_4$ is a punctured disc with center $z=-3$, radius $5$ and the pole at $z=2$ at the boundary of the disc. It admits for the fraction with pole at $z=-3$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.

• The region $D_5$ contains all points outside the disc with center $z=-3$ and radius $5$. It admits for both fractions a representation as principal part of a Laurent series.

The expansion of $f$ as Laurent series in $D_4$ is

\begin{align*} f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\ &= \frac{6}{z+3} +\frac{4}{(z+3)-5}\\ &= \frac{6}{z+3} -\frac{4}{5}\cdot\frac{1}{1-\frac{z+3}{5}}\\ &= \frac{6}{z+3} -\frac{4}{5}\sum_{n=0}^\infty\frac{1}{5^{n}}(z+3)^{n}\\ \end{align*}

The expansion of $f$ as Laurent series in $D_5$ can be calculated similarly.

Answer 2

HINT: let $g(z):=\frac1{z-1}$. Then there are two distinct Laurent expansions of $g$ around $z=-3$ depending of the considered domain.

Observe that

$$g(z)=\frac1{z-1}=\frac1{(z+3)-4}=\begin{cases}\frac1{z+3}\cdot\frac1{1-\frac4{z+3}}=\frac1{z+3}\sum_{k=0}^\infty\left(\frac4{z+3}\right)^k,& |z+3|>4\\ -\frac14\cdot\frac1{1-\frac{z+3}4}=-\frac14\sum_{k=0}^\infty\left(\frac{z+3}4\right)^k,&|z+3|<4\end{cases}$$