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Take the function $f(z)=(z^2+3z+2)e^\frac{1}{z+1}$

We want to expand this into its Laurent series about $z_0$=-1.

Alright, so I'm a little confused. This converges everywhere but -1, which throws me off. My first thought is to just start expanding it using properties that I know. So, I think we have the following:

$e^z=1+z+\frac{1}{2!}z^2+\frac{1}{3!}z^3+...$

$\frac{1}{1-(-z)}=1-z+z^2-z^3+z^4-...$

But I'm not sure how to expand the term $e^\frac{1}{z+1}$. It seems like a composition of expansions. Also, can I use a substitution to account for the point -1? This seems like it's going to turn out very messily. I'd appreciate any guidance.

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Hint. You may just write, for $z\neq-1$, $$e^{\large\frac1{z+1}}=1+\frac1{(z+1)}+\frac1{2!(z+1)^2}+\cdots+\frac1{n!(z+1)^n}+\cdots $$ and $$ (z^2+3z+2)=(z+1)^2+(z+1) $$ then expand $$ \left((z+1)^2+(z+1)\right)\times\left(1+\frac1{(z+1)}+\frac1{2!(z+1)^2}+\cdots+\frac1{n!(z+1)^n}+\cdots\right) $$ Can you take it from here?


Set $X=z+1$, you get $$ \begin{align} (X^2+X)\sum_{0}^{\infty}\frac1{n!}\frac1{X^n}&=\sum_{0}^{\infty}\frac1{n!}\frac1{X^{n-2}}+\sum_{0}^{\infty}\frac1{n!}\frac1{X^{n-1}}\\\\ &=\sum_{-2}^{\infty}\frac1{(n+2)!}\frac1{X^{n}}+\sum_{-1}^{\infty}\frac1{(n+1)!}\frac1{X^{n}}\\\\ &=\frac1{X^{-2}}+\sum_{-1}^{\infty}\left(\frac1{(n+2)!}+\frac1{(n+1)!}\right)\frac1{X^{n}}\\\\ &=\frac1{X^{-2}}+\sum_{-1}^{\infty}\frac1{(n+1)!}\left(\frac1{(n+2)}+1\right)\frac1{X^{n}}\\\\ &=\frac1{X^{-2}}+\sum_{-1}^{\infty}\frac{n+3}{(n+2)!}\frac1{X^{n}}\\\\ &=\sum_{-2}^{\infty}\frac{n+3}{(n+2)!}\frac1{X^{n}} \end{align} $$ Then the Laurent series expansion of your function is

$$ (z^2+3z+2)e^{\large \frac{1}{z+1}}=\sum_{-2}^{\infty}\frac{n+3}{(n+2)!}\frac1{(z+1)^{n}},\quad z\neq-1. $$

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  • $\begingroup$ Thank you so much for your assistance! I think I'm missing a fundamental piece of knowledge - how does the term $z_0=-1$ actually come into play using this notation? $\endgroup$
    – Taylor
    May 1, 2015 at 3:37
  • $\begingroup$ @Taylor This way:$z-z_0=z-(-1)=z+1$. Thanks! $\endgroup$ May 1, 2015 at 3:39
  • $\begingroup$ Oh that makes sense! So all I have to do is foil out the product, and that produces a Laurent series expansion about $z_0=-1$? Thanks so much. $\endgroup$
    – Taylor
    May 1, 2015 at 3:54
  • $\begingroup$ @Taylor You are welcome. Maybe, if you put $X=\frac1{z+1}$, things will be clearer to finish the computation. $\endgroup$ May 1, 2015 at 3:56
  • $\begingroup$ If you don't mind me asking one more clarifying question, how is this different from expanding the series $e^{\frac{1}{z}}$ about the point $z_0=-1$? Wouldn't that result in the same expansion? This makes me think the expansion should be $e^{\frac{1}{z+1}}=1+\frac{1}{z}+...$, which looks like the expansion of $e^{\frac{1}{z}}$ at 0. Can you enlighten me? @Olivier $\endgroup$
    – Taylor
    May 2, 2015 at 4:00

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