I need to find Laurent expansion for $$\frac{e^{\frac{1}{z-1}}}{z(z+1)}$$ about $z=1$ for $1 \lt |z-1| \lt 2$.

I've tried to solve this as it was suggested in the book: find Laurent expansion for $f_1(z) = \frac{1}{z+1}$ about $z=1$ for $|z-1| \lt 2$ and Taylor expansion for $f_2(z) = \frac{e^{\frac{1}{z-1}}}{z}$ for $|z-1| > 1$ (this function is analytic in this area, so Taylor series is equal to Laurent series) and then multiply them.

I get $\frac{1}{z+1} = \sum_{n=-\infty}^{-1}(-1)^{n-1}2^{-n-1}(z-1)^n$, $\frac{1}{z} = \sum_{n=0}^{\infty}(-1)^n(z-1)^n$ and $e^{\frac{1}{z-1}} = 1+ \sum_{n=1}^{\infty}\frac{1}{n!(z-1)^n}$.

But I don't understand how can I find product of these series?

Thanks for your help.


Hint: With $z=u+1$ $$\dfrac{1}{z(1+z)}=\dfrac{1}{1+u}-\dfrac{\frac12}{1+\frac{u}{2}}=\sum_{n=0}(-u)^n-\dfrac12\sum_{n=0}(-\frac{u}{2})^n=\dfrac{1}{2}-\dfrac{3}{4}u+\dfrac{7}{8}u^2-\dfrac{15}{16}u^3+\cdots$$ now find the product of this series and the series of $e^{\frac{1}{u}}$ and substitute $u=z-1$.

  • $\begingroup$ Why do we make the substitution z=u+1? $\endgroup$ – blargen Apr 12 '19 at 17:41
  • $\begingroup$ @blargen Just for simplifying the calculations. $\endgroup$ – Nosrati Apr 12 '19 at 18:35

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