Contour integrals in complex analysis that don't use a closed contour - do we have path independence? I've noticed that the vast majority of integration problems that I work on in complex analysis are on closed contours, using the Residue Theorem.  (If the contour is not closed, we usually close it with a big circle or semi-circle or box.)
Does anyone ever integrate from one starting point to a different ending point in complex analysis?  And, integrating on this non-closed path, would we have path-independence?
Lately, I've been trying to make as close of a connection as possible between complex and real analysis.  My first mistake was in claiming that holomorphic functions are conservative, trying to agree with the Cauchy-Goursat integral theorem in a simply connected domain.  It turns out that the conjugate is conservative, but not the function itself.
If we get path-independence in complex analysis, then it should follow that this is not a consequence of the integrand being conservative (it is not necessarily equivalent to a gradient field on $R^2$), but from something else.  What is this something else? 
Thanks,
 A: Consider a holomorphic function $f$ in a simply connected domain $\Omega$, where $a, b$ are two points in $\Omega$. Since $f$ is holomorphic, it has a primitive $F$ with $F' = f$. Then
$$
\int_\gamma f(z) dz = F(b) - F(a)
$$
for any curve $\gamma$ from $a$ to $b$ completely contained within $\Omega$. That is, if $f$ is holomorphic, then line integrals of $f$ are path-independent.
We can prove path independence rather quickly. Suppose $\gamma_1$ and $\gamma_2$ are two paths from $a$ to $b$ in $\Omega$. Then $\Gamma = \gamma_1 \cup \overline{\gamma_2}$ (meaning first go along $\gamma_1$ from $a$ to $b$, and then go along $\gamma_2$ backwards from $b$ back to $a$) is a closed loop. Thus
$$ \int_\Gamma f(z) dz = \int_{\gamma_1} f(z)dz - \int_{\gamma_2} f(z)dz = 0$$
from Cauchy's Theorem. Rearranging, we have
$$ \int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z)dz$$
which precisely says that the line integral is path-independent.
So we've seen that $f$ being holomorphic is sufficient to guarantee path-independence. Let us now show that it is also necessary.

Claim: Let $\Omega$ be a simply-connected domain and let $f$ be a continuous function function on $\Omega$. If the integral of $f$ along any closed path in $\Omega$ is $0$, then $f$ has a holomorphic primitive $F$ with $F' = f$.

To prove this, we pick an arbitrary point $a$ within $\Omega$ and define
$$
F(w) = \int_a^w f(z)dz,
$$
where the integral is taken along any path from $a$ to $w$. This is well-defined since if $\gamma_1$ and $\gamma_2$ are two such paths, then
$$ \int_{\gamma_1} f = \int_{\gamma_2} f$$
wince $\gamma_1 \overline{\gamma_2}$ is a closed loop, just like above. Let us explicitly show that $F' = f$. So consider
$$ \frac{F(z+h) - F(z)}{h} = \frac{1}{h} \int_z^{z+h} f(\zeta)d\zeta.$$
Since the integral is path independent, I'm just going to assume that we are looking at the path segment from $z$ to $z+h$. As $f$ is continuous at $z$, we have $f(\zeta) = f(z) + \phi(\zeta)$ for some function $\phi$ such that $\phi(\zeta) \to 0$ as $\zeta \to z$. Then
$$ \frac{1}{h} \int_z^{z+h} f(\zeta)d\zeta = \frac{1}{h} \int_z^{z+h} f(z) d\zeta + \frac{1}{h} \int_z^{z + h} \phi(\zeta) d\zeta.$$
The first term is precisely $f(z)$. The second is bounded above by
$$ \frac{1}{|h|} |h| \max \phi(\zeta)$$
which goes to $0$ as $h \to 0$ (since $\zeta \to z$ in that case). We get the derivative by letting $h \to 0$, and thus we see that $F' = f$ and $F$ is a primitive for $f$. $\spadesuit$
To be one-time complex-differentiable is to be infinitely differentiable. So to have a holomorphic primitive is the same as being holomorphic. So we can conclude that $f$ being holomorphic is equivalent to $f$ having path-independent line integrals within a simply connected domain.
