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I am trying to find the Laurent series and region of convergence $\frac{z}{(z+2)(z+1)}$ at $z=-2$. I found that

$$ \frac{z}{(z+2)(z+1)}=\frac{2}{z+2}+\sum^\infty_{n=0}(z+2)^n $$

But I am confused because usually when I am asked to find the radius of convergence of Laurent series I just calculate the radius for the series but for the function above we have a series plus the term $\frac{2}{z+2}$ so do I just find the radius of convergence of the series and ignore the term $\frac{2}{z+2}$?

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  • $\begingroup$ $\left\{\,z\,\,\, {\Large \mid}\,\,\, \left\vert\,z + 2\,\right\vert < 1 \wedge z \not= 2\,\right\}$. $\endgroup$ – Felix Marin May 26 '17 at 20:16
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Here we have to find two radii: the inner radius and the outer radius. Inner radius: the function $\frac{2}{z+2}$ is not defined at $z=-2$, so we need that $|z+2|>0$. Outer radius: the series $\sum^\infty_{n=0}(z+2)^n$ is convergent for $|z+2|<1$. Putting all together we have that the region of convergence is $0<|z+2|<1$.

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  • $\begingroup$ $0<|z+2|<1$. But that is not what I am asking. My question do I just ignore the terms $\frac{2}{2+z}$ (the terms outside the sum) when find the region of convergence. $\endgroup$ – gbd May 26 '17 at 7:38
  • $\begingroup$ No. That term give you the condition $|z+2|>0$. Without that term the region of convergence would be $|z+2|<3$ ( $z=-2$ is allowed). $\endgroup$ – Robert Z May 26 '17 at 7:40
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If you set, as $z \to -2$, $$ u=z+2,\quad u \to 0, $$ then you get, for $|u|<1$ and $u\neq0$, $$ \frac{z}{(z+2)(z+1)}=\frac{u-2}{u(u-1)}=\frac{2}{u}+\frac{1}{1-u}=\frac{2}{u}+\sum_{n=0}^\infty u^n=\frac{2}{z+2}+\sum^\infty_{n=0}(z+2)^n $$ the latter Laurent series then exists for $0<|z+2|<1$.

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If you have a laurent series $$ \sum_{k=1}^\infty a_{-k}(z-z_0)^{-k}+\sum_{k=0}^\infty a_k(z-z_0)^k $$ then you have to compute $$ r=\limsup_{k\to\infty} \sqrt[k]{|a_{-k}|} $$ and $$ R=\frac{1}{\limsup_{k\to\infty} \sqrt[k]{|a_{k}|}}. $$ Then your laurent series is defined for $r<|z-z_0|<R$.

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