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

Now let [itex] C [/itex] be the unit circle, oriented counterclockwise. Let's also say that 1 is the initial and final point of [itex] C [/itex]. Then by direct computation, we know that [tex] \int_C \frac{1}{z} dz = 2 \pi i .[/tex] Now, I think I understand that [itex] \log z [/itex] is not quite an antiderivative of [itex] \frac{1}{z} [/itex], 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 [tex] \int_C \frac{1}{z} dz = \log(1) - \log(1). [/tex] But on the other hand, it kind of looks like the formula still holds, if we understand that the final value of [itex] \log(1) [/itex] is [itex] 2\pi i [/itex] 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 [itex] f [/itex] has an antiderivative [itex] F [/itex] that is multiple-valued, like in my example above, can we still say that [tex] \int_C f(z) dz = F(z_1) - F(z_0), [/tex] as long as we choose the "appropriate" value of [itex] F(z_1) [/itex] and [itex] F(z_0) [/itex]? And if so, how could one go about proving this?

Thanks for reading all of that.