0
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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 $f$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.