Complex Integrals, Antiderivatives, Logarithms

[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.

 

[micromass]

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

 

[WWGD]

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.

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.