Let $f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}$, find the Laurent series about $z=0$.

On the region $0<|z|<2$, I get


then $\cfrac{1}{z^2(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$

And I get confused here, is it ok if I let


Or I have to express $e^{-3z}$ as its Taylor series and the multiplicate?

Thank you

  • $\begingroup$ That is exactly it. I think you have a typo, though. between the third and fourth lines, the left hand side changes, the right hand side does not. $\endgroup$ – Alex S Jun 27 '15 at 4:53
  • $\begingroup$ Thank you I've edited it. So, the answer will be $f(z)=e^{-3z}\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$ ? $\endgroup$ – Frank Jun 27 '15 at 5:11
  • $\begingroup$ It's not OK to leave it like that, as that isn't a Laurent series. Whatever the answer is, it will be the product of the Taylor and Laurent series, but I can't help but feel there must be a better way than just multiplying them! $\endgroup$ – Theo Bendit Jun 27 '15 at 5:18
  • $\begingroup$ I had a play around with multiplying the series manually, and I found that the existence of a closed form for the coefficients is equivalent to the existence of a closed form for a sum of the form $\sum_{k=0}^n \frac{x^k}{k!}$. I'm sceptical that such a closed form exists, but maybe someone else could shine a light here? $\endgroup$ – Theo Bendit Jun 27 '15 at 5:48
  • $\begingroup$ The expansion is doable for sure but the problem is to find the explicit form for the coefficient. $\endgroup$ – Claude Leibovici Jun 27 '15 at 5:55

You have $$f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}=\Big(\sum_{m=0}^\infty \frac{(-3)^m }{m!}z^m\Big)\times \Big(\sum_{n=0}^\infty \frac{n}{2^{n+1}}z^{n-3}\Big)=\sum_{i=-2}^\infty c_i z^i$$ and you look for the general expression of the $c_i$'s.

For better legibility, we shall write $$\sum_{i=-2}^\infty c_i z^i=\Big(\sum_{m=0}^\infty a_m z^m\Big)\times \Big(\sum_{n=0}^\infty b_n z^{n-3}\Big)$$

In order to have a degree $(k-3)$, you must add several terms the product of whicb making $z^k$ and then you should arrive to something like $$c_{k+3}=\sum_{m=0}^k a_m \,b_{k-m}$$

  • $\begingroup$ Thank you very much, I had the same idea but i wanted to know if ther was another way to do so. $\endgroup$ – Frank Jun 29 '15 at 23:29

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