Here is the solution to the problem (I need some clarification):

After partial fraction decomposition we have $\dfrac{1}{z(1+z)} = \dfrac{1}{z} - \dfrac{1}{1+z}$

But since we're only interested by the region $0< |z| < 1$, we can rewrite $\dfrac{1}{1+z} = \dfrac{1}{1-(-z)} = \sum_{k≥0} (-z)^k$

Therefore $f(z) = \dfrac{1}{z} -  \sum_{k≥0} (-z)^k$ is the Laurent series.

The problem I have is that we defined a Laurent series as a series of the form $\sum_{k=-\infty}^{\infty}a_k (z-c)^k$ And $\dfrac{1}{z} -  \sum_{k≥0} (-z)^k$ is clearly not of that form.


Yes, it is of that form. With $a_j = 0$ for $j < -1$, and $a_k = (-1)^{k+1}$ for $k \geq -1$. And, of course, $c = 0$.

  • $\begingroup$ :-P Not going to waste my time posting the same as your answer, so $+1$. $\endgroup$ – Simply Beautiful Art Jan 8 '17 at 13:28

Note that $\frac{1}{z(z+1)}$ has a simple pole at $z=0$ so the Laurent's expansion has $a_{-n}=0$ for all $n\neq 1$, $n\in \mathbb N$

  • $\begingroup$ Taylor series do not have negative exponent terms. Laurent series do. $\endgroup$ – Arthur Jan 8 '17 at 14:37
  • $\begingroup$ @Arthur...thanks for pointing it. I have edited it. $\endgroup$ – Nitin Uniyal Jan 8 '17 at 16:16

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