# Laurent series of $f(t)$

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?

• What do you mean by "fo"? – Ron Gordon Dec 22 '13 at 9:36
• sorry, i have editied it, it didnt paste properly :) – R.A Dec 22 '13 at 10:00
• 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? – Greg Martin Dec 22 '13 at 10:26
• 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? – R.A Dec 22 '13 at 10:29

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 $f$.