Cauchy integral formula

Suppose I have a function {f} which is holomorphic on an open set {U} in {{\mathbb C}} (you should be used to this setup by now). Pick some point {a\in U}. How what about the function {f(z)/(z-a)}? It’s easy to check that it’s holomorphic on {U-\{a\}}. 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 {\gamma} denote a loop which goes around {a}, 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 {a}. 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 {\gamma} (the loop we started with), a straight line segment which we’ll call {\sigma}. A small circle going around {a} which we’ll call {\beta}, and the same segment {\sigma}, but traversed in the opposite direction. The whole thing is the boundary of the shaded region. Notice that if {\gamma} is traversed counter-clockwise, then {\beta} is traversed clockwise. Integrating over the entire curve yields zero. Since we would go across {\sigma} once forwards and once backwards, it must contribute zero. Thus, it follows that

\displaystyle 0=\int_\gamma\frac{f(z)}{z-a}dz+\int_\beta \frac{f(z)}{z-a}dz.

But this is where {\beta} is traversed clockwise. If we went counterclockwise, the two integrals (around {\gamma} and around {\beta} would be equal. Moreover, I never said how small a circle {\beta} 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 {f}, {a}, and {\gamma} be as above. Then

\displaystyle f(a)=\displaystyle\int_\gamma\frac{f(z)}{z-a}dz.

Proof: Like we said, it doesn’t matter which loop we take, so take a loop {\beta_r} which is a counter-clockwise circle of radius {r} about {a}, and compute

\displaystyle \int_{\beta_r}\frac{f(z)}{z-a}dz-f(a).

Let {z} be a function of {\theta} given by {z=a+r\cdot e^{i\theta}}. Then {dz=re^{i\theta}\cdot id\theta=(z-a)id\theta}

\displaystyle \begin{array}{rcl} \frac1{2\pi i}\displaystyle\int_{\beta_r}\frac{f(z)}{z-a}dz-f(a) &=& \frac1{2\pi i}\displaystyle\int_0^{2\pi}f(z)\cdot id\theta-f(a)\\ &=& \frac1{2\pi }\displaystyle\int_0^{2\pi}f(z)\cdot d\theta-f(a) \end{array}

The next step is to move {f(a)} into the integral. To do so, we need to multiply it by {2\pi} because of the {1/(2\pi)} outside the integral. But we also need to divide by {2\pi} because we’re integrating a constant from {0} to {2\pi}. In other words, the fraction on the outside “makes up” for the integral, and we can pull {f(a)} straight through. This gives us

\displaystyle \frac1{2\pi}\int_0^{2\pi}\left(f(z)-f(a)\right)d\theta.

Now how big can {f(z)-f(a)} get? I don’t have the slightest clue, but I can tell you that for the values {z} we pick on the circle of radius {r} about {a}, {f(z)-f(a)} cannot be arbitrarily large in absolute value. This is simply because the circle is a compact set, and every real function (such as {|f(z)-f(a)|} attains its maximum on a compact set. Let {M_r} be the maximum on magnitude of {f(z)-f(a)} attained on the circle of radius {r}. Then

\displaystyle \begin{array}{rcl} \left|\frac1{2\pi}\int_0^{2\pi}\left(f(z)-f(a)\right)d\theta\right| &\le& \frac1{2\pi}\int_0^{2\pi}\left|f(z)-f(a)\right|d\theta\\ &\le& \frac1{2\pi}\int_0^{2\pi}M_r d\theta\\ &=& M_r \end{array}

As we mentioned earlier, it doesn’t matter what {r} we pick, and since {f} is continuous, very small choices in {r} yield very small changes between {f(z)} and {f(a)}, and so we can make {M_r} as small as we like. Altogether this says that the magnitude of

\displaystyle \frac{1}{2\pi i}\displaystyle\int_\gamma\frac{f(z)}{z-a}dz-f(a)

is zero, meaning the quantity itself is zero, which proves the theoerm. \Box


4 Responses to Cauchy integral formula

  1. Pingback: Cauchy integral formula again « Andy Soffer

  2. Pingback: Cauchy estimates « Andy Soffer

  3. Pingback: Laurent series « Andy Soffer

  4. Pingback: Meromorphic functions and residues « Andy Soffer

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s