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Find the Laurent expansion of $\frac{z}{(z+1)(z+2)}$ about the singularity $z=-2$. Specify the region of convergence and the nature of singularity at $z = -2$.

The Laurent expansion I get is

$1+(z+2)+(z+2)^2+ \ldots + \frac{2}{z+2}$

The singularity is a simple pole.

But how to find the Region of Convergence.

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The Laurent series can be obtained as

$$\frac{z}{(z+1)(z+2)} = \frac{1}{z+2}\frac{z+2 - 2}{z+2 - 1} \\ = \left(\frac{2}{z+2} - 1\right)\frac{1}{1 - (z+2)} \\ = \left(\frac{2}{z+2} - 1\right)\sum_{n=0}^\infty (z+2)^n$$

and the geometric series on the RHS converges when $|z+2| < 1$.

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The function \begin{align*} f(z)&=\frac{z}{(z+1)(z+2)}\\ &=\frac{2}{z+2}-\frac{1}{z+1}\\ \end{align*} is to expand around the center $z=-2$. Since there are simple poles at $z=-1$ and $z=-2$ we have to distinguish two regions of convergence \begin{align*} D_1:&\quad 0<|z+2|<1\\ D_2:&\quad |z+2|>1 \end{align*}

  • The first region $D_1$ is a punctured disc with center $z=-2$, radius $1$ and the pole $-1$ at the boundary of the disc.

    In the interior we have a representation of the fractions with pole at $z=-2$ as principal part of a Laurent series at $z=-2$, while the fraction with pole at $z=-1$ admits a representation as power series.

  • The other region $D_2$ containing all points outside the closure of $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=-2$.

Expansion in $D_1$: \begin{align*} f(z)&=\frac{1}{1-(z+2)}+\frac{2}{z+2}\\ &=\frac{2}{z+2}+\sum_{n=0}^\infty(z+2)^n \end{align*}

Expansion in $D_2$:

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

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  • $\begingroup$ +1. Nice job. I feel that MSE is in a summer vacation !!!. $\endgroup$ – Felix Marin Jul 31 '17 at 4:35
  • $\begingroup$ @FelixMarin: Thanks Felix. It seems you're right. :-) $\endgroup$ – Markus Scheuer Jul 31 '17 at 6:02
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Consider the Taylor series expansion of $\frac{z}{(z+1)(z+2)}-\frac{2}{z+2}=\frac{-1}{z+1}$. Its radius of convergence around $z=-2$, and therefore the radius of convergence of your Laurent series, is simply the distance from $-2$ to the nearest non-differentiable point, i.e. $z=-1$.

So the radius of convergence is 1.

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