Suppose $f$ is analytic on the domain $D = \{z \big| |z|>1\}$ in $\mathbb{C}$. I know that if the line integral of $f$ along any closed path is 0 then $f$ has an anti-derivative. Is this also a necessary condition.
Necessary condition for existence of anti-derivative
3
$\begingroup$
complex-analysis
-
1How would you compute the line integral if you knew that, say, $F(z)$ were an anti-derivative? – 2010-12-17
-
0I'm rather confused by your statement. For any analytic function $f$, by the [Cauchy-Goursat theorem](http://en.wikipedia.org/wiki/Cauchy-Goursat_theorem), its integral along any closed path is 0. Are you sure you are correctly stating what you want? – 2010-12-17
-
0The domain I am looking at is not simply connected so I dont believe Cauchy-Goursat applies. – 2010-12-17
-
0Oh, oops, read the inequality the other way. You really shouldn't use $D$ for a domain that is not a disk :p (Though steps used in the *proof* of Cauchy-Goursat will also give you what you want, like Jonas' answer below.) – 2010-12-17
1 Answers
5
Yes. If $F'=f$ and $\gamma:[a,b]\to D$ is a curve, then $\int_\gamma f = F(\gamma(b))-F(\gamma(a))$. For example, see Theorem 1.18 on page 65 of John B. Conway's Functions of one complex variable.