Prove that for any Laurent series $f(t)$ one has $\operatorname{Res}\{f'\} = 0$?

I know for a Laurent series of a complex function f is a representation of that function as a power series which includes terms of negative degree. So for example the sum goes from -inf to inf.

I am really stuck on this question?

  • 1
    $\begingroup$ What do you mean by "fo"? $\endgroup$ – Ron Gordon Dec 22 '13 at 9:36
  • $\begingroup$ sorry, i have editied it, it didnt paste properly :) $\endgroup$ – R.A Dec 22 '13 at 10:00
  • $\begingroup$ Do you know what the definition of the residue is? Do you know the relationship between the residue of a function and its Laurent series? $\endgroup$ – Greg Martin Dec 22 '13 at 10:26
  • $\begingroup$ I have googled the defintion, and i know residues can be calculated by finding Laurent series expansions, and one can define the residue as the coefficient a-1 of a Laurent series. I just dont know how todo it? $\endgroup$ – R.A Dec 22 '13 at 10:29

There are a few ways to see this:

1) The derivative of the Laurent series $\sum_{i=n}^\infty a_i z^i$ is $\sum_{i=n-1,~i\neq -1}^\infty (i+1)a_{i+1}z^{i}$, the coefficient of $z^{-1}$ is $0$.

2) If we go by the first definition on Wikipedia: $Res_a(f)$ is the unique complex number such that $f(z)-\frac{R}{z-a}$ has an analytic antiderivative in a punctured disk around $a$. Now $f'$ clearly has an analytic antideriavative, namely $f$.

  • $\begingroup$ Is that all it was? I honestly thought it wasn't that simple :/ lol $\endgroup$ – R.A Dec 22 '13 at 10:42

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