 Suppose our function is analytic in a connected domain R. Koshy Gorsah tells us that the integral is zero for all simple closed curves in R. But what if C isn't closed? So suppose C1 is a path between Z0 and Z1. If C2 is a different path between Z0 and Z1, then C1 minus C2, that's going along path C2 backwards, forms a closed curve. So the difference of the integrals is zero. Consequently, the integral along C1 is equal to the integral along C2. And this gives us an important theorem. If our function is analytic in a connected region, then the value of the definite integral is independent of the actual path between Z0 and Z1. Now remember, we've been assuming that our integral is path independent all along, and this finally gives us what we need. It has to be analytic in a connected region. Now as an advanced math course, we should be a little bit careful here. While the theorem is true, you should recognize that there is some hand-waving. Notice that we got rid of the requirement that we have a simple curve. So what if our curve crosses itself? And even if our first curve doesn't, what if C2 crosses C1? So C1 minus C2 isn't simple. And can we always find a closed path that includes our curve C1? The first two questions are actually relatively easy to address. The third is rather more complicated. This hand-waving aside, suppose F of Z is analytic in a connected region R. Let gamma over some interval parameterize a simple path, C, from Z0 to Z1. Then the contour integral will be... And since this is now a single variable calculus problem, we can use the fundamental theorem of single variable calculus to find. Consequently, if F is analytic in a connected region R, then the integral is pretty much what we'd expect it to be. So we can now evaluate an integral like the integral from 1 to I of cosine Z, DZ. Since cosine is analytic everywhere, the fundamental theorem applies, so we have... Note that our connected region doesn't have to be all of the complex plane. All we need to do is find some connected region that avoids the bad places where F of Z is not analytic. So suppose F of Z is analytic everywhere, but at the point shown. If possible, find a connected region R that includes Z equals negative I, Z equals I, where F of Z is analytic in R. If not, explain why not. And we can avoid the bad places by taking the region like this one. Now, a detailed description of the region is generally unnecessary as long as the region actually exists. Well, remember, it's your grade. So suppose F of Z is analytic everywhere, but along the ray shown. Again, let's try to find a connected region that includes Z equals negative I and Z equals I, and we can do that if we take a C-shaped region. What if we have an entire line of non-analyticity? Unfortunately, the line of non-analyticity splits our complex plane into two parts where negative I is in one part and I is in the other. So it's not possible to find a connected region that excludes the line, but includes both points. So let's find the integral of 1 over Z from 1 to I. Now, 1 over Z isn't analytic at Z equals 0, but as long as we avoid the bad place, we can apply the fundamental theorem. Since there is a connected region R that includes 1 and I, where 1 over Z is analytic, we have...