 Suppose our path C is closed, in other words, we get back to our starting point. We can break this path into two parts, say C1 going from some starting point to some midpoint, and C2 going from the midpoint back to the starting point. By the additivity of the integrals, the integral around the entire path is the integral around C1 plus the integral through C2. We'll assume, for now, that the value of the integral only depends on its endpoints, so... Then this first integral is some function at C1 minus the same function at C0. And the second integral will be some function at C0 minus some function at C1. And if we simplify, we get zero. And this suggests a theorem, let C be a simple closed path. Then the contour integral along Z for f of Z is going to be zero. And this is such an amazing and powerful result that we have to wonder whether this is actually true. So let's find the value of these contour integrals around closed paths for some different functions. So let's see if the unit circle centered at Z equals zero. Let's evaluate the contour integral around C. So remember, we can parameterize the unit circle as cosine t plus i sine t, or even more conveniently in exponential form, e to power i t. And so our integral will become... And if we can evaluate this integral using a u substitution with u equal e to the i t and du equal i e to the i t dt. And if we do that, we find... Which is consistent with our expectations. Well, let's try sine z. So as before, we can let z equal e to the i t and then a u substitution to obtain. And if we have the integral of one over z, our work suggests we'll get a value of zero. But let's check it out. So again, we'll let z equal e to the i t, and so we get. And we have an integral not equal to zero. So let's think about that. When c is the unit circle centered around the origin, we found a couple of integrals were zero and one was not. The obvious difference is that one over z is undefined on the region inside c. So is that what's making all the difference? Let's take a look. So we might try something like one over z squared. So as before, we'll parameterize z as e to the i t, dz is i e to the i t, and so we find... And so we get something that isn't not equal to zero. And in fact, it's a little bit worse than that. Let's consider something like the conjugate of z, which is defined for all points inside the unit circle. So as before, we'll let z equal e to the i t and find. And even though the conjugate is defined everywhere inside the unit circle, we still get a non-zero value for the integral. So clearly, there's something very strange is going on. The critical observation here is that the conjugate of z is not differentiable anywhere, and one over z and one over z squared are not differentiable at points inside the circle. Meanwhile, z and sine z are differentiable. Well, actually even better, they're analytic at every point inside the circle. And this suggests the following. Let z be a simple closed path, and suppose f of z is analytic in a region that includes the interior of z. Then the contour integral around z is equal to zero. This is known as the Cauchy-Gorsuch theorem, and it's one of the most important theorems in complex analysis.