# Complex line integral independent of path

Could you please verify my solution?

Let $f$ be holomorphic on an open set $U$ which is the interior of a disc or a rectangle. Let $p,q \in U$. Let $\gamma _j : [a,b] \to U, j = 1,2,$ be $C^1$ curves such that $\gamma _j(a) = p, \gamma _j(b)=q, j =1,2$. Show that $$\oint _{\gamma _1} \ fdz = \oint _{\gamma _2} \ fdz$$ Function Theory... 2.2

Solution: consider $\gamma := \gamma_1 - \gamma_2$. This is clearly closed by definition (and is piecewise $C^1$), so $$\oint_{\gamma} \ fdz = 0 = \oint_{\gamma_1}\ fdz - \oint_{\gamma_2}\ fdz$$

as desired.

Is this sufficient? I got the idea from (what seemed to be) a related question. Having just downloaded my first text on Diffential Forms, I can't fully grasp what is said there, but I took a stab at a similar argument.

This exercise appears to be saying that the line integral is independent of path. There is mention of the complex case on the Wikipedia page, but their approach was to convert the integral to something real valued. I would also be interested in a solution that uses that technique.

Thanks

-
I can't help but wonder whether the interesting formation "ingested" was produced by a spell-checker gone astray or by association with a widespread pronunciation of "interested" with an affricate that sounds not too unlike "ingested" (but with the emphasis on a different syllable)? –  joriki Feb 14 '12 at 23:23
Yes, @joriki, this was typed using iOS! Do you have any thoughts on the rest? :) –  The Chaz 2.0 Feb 15 '12 at 1:29

You should probably mention that since the domain is simply connected, Cauchy's theorem says that $\int_\gamma f \ dz = 0$.