Complex Integrals, Antiderivatives, Logarithms

1. Nov 1, 2015

Local

I've been teaching myself a little bit of Complex Variables this semester, and I had a question concerning complex integrals.

If I understand correctly, then if a function $f$ has an antiderivative $F$, then the line integral $\int_C f(z) dz$ is path independent and always evaluates to $F(z_1) - F(z_0)$, where $z_1$ and $z_0$ are the final and initial points of the contour $C$.

Now let $C$ be the unit circle, oriented counterclockwise. Let's also say that 1 is the initial and final point of $C$. Then by direct computation, we know that $$\int_C \frac{1}{z} dz = 2 \pi i .$$ Now, I think I understand that $\log z$ is not quite an antiderivative of $\frac{1}{z}$, because there is no way to define a single-valued branch of the complex logarithm over a region containing the unit circle in such a way that it will be differentiable at every point of the circle. So we can't quite say that $$\int_C \frac{1}{z} dz = \log(1) - \log(1).$$ But on the other hand, it kind of looks like the formula still holds, if we understand that the final value of $\log(1)$ is $2\pi i$ more than the initial value, because we have gone around full circle.

At this point, I'm not too sure what my question is. I guess my question is: If a function $f$ has an antiderivative $F$ that is multiple-valued, like in my example above, can we still say that $$\int_C f(z) dz = F(z_1) - F(z_0),$$ as long as we choose the "appropriate" value of $F(z_1)$ and $F(z_0)$? And if so, how could one go about proving this?

Thanks for reading all of that.

2. Nov 2, 2015

micromass

Staff Emeritus
You will see the beautiful solution of this when studying the Riemann surface of a function.

3. Nov 2, 2015

WWGD

An actual antiderivative exists within a branch cut. EDIT: the antiderivative can be translated by
integer multiples of $2\pi i$ to get a different antiderivative in a different branch cut.

Last edited: Nov 2, 2015