Cauchy integral formula
July 16, 2012 4 Comments
Suppose I have a function which is holomorphic on an open set in (you should be used to this setup by now). Pick some point . How what about the function ? It’s easy to check that it’s holomorphic on . You can do this by proving the applications of the multiplication rule and chain rule with one of the example functions we saw here.
Now let denote a loop which goes around , such as the one to the left. The Cauchy integral theorem says that the integral is zero, so long as the loop has an interior, but this one doesn’t (it’s missing the point . But we can be tricky and make a different loop composed of four parts. Take a look at the picture to the right.
The four parts are (the loop we started with), a straight line segment which we’ll call . A small circle going around which we’ll call , and the same segment , but traversed in the opposite direction. The whole thing is the boundary of the shaded region. Notice that if is traversed counter-clockwise, then is traversed clockwise. Integrating over the entire curve yields zero. Since we would go across once forwards and once backwards, it must contribute zero. Thus, it follows that
But this is where is traversed clockwise. If we went counterclockwise, the two integrals (around and around would be equal. Moreover, I never said how small a circle was. Indeed, it can be as small as we like. We’ll use this to prove the following nice result:
Theorem 1 (Cauchy integral formula) Let , , and be as above. Then
Proof: Like we said, it doesn’t matter which loop we take, so take a loop which is a counter-clockwise circle of radius about , and compute
Let be a function of given by . Then
The next step is to move into the integral. To do so, we need to multiply it by because of the outside the integral. But we also need to divide by because we’re integrating a constant from to . In other words, the fraction on the outside “makes up” for the integral, and we can pull straight through. This gives us
Now how big can get? I don’t have the slightest clue, but I can tell you that for the values we pick on the circle of radius about , cannot be arbitrarily large in absolute value. This is simply because the circle is a compact set, and every real function (such as attains its maximum on a compact set. Let be the maximum on magnitude of attained on the circle of radius . Then
As we mentioned earlier, it doesn’t matter what we pick, and since is continuous, very small choices in yield very small changes between and , and so we can make as small as we like. Altogether this says that the magnitude of
is zero, meaning the quantity itself is zero, which proves the theoerm.