Cauchy integral formula again

We’ve already proved the Cauchy integral formula:

Theorem 1 (Cauchy integral formula) For {f\in{\mathcal H}(U)}, {z\in U}, and any counterclockwise circle {\gamma} in {U} about {z} (or any loop homotopy equivalent to it in {U-\{z\}}),

\displaystyle f(z)=\frac1{2\pi i}\int_\gamma \frac{f(w)}{w-z}dw.

It’s really quite amazing. Thinking about it slightly more algebraically, consider the operator

\displaystyle F\mapsto\left(z\mapsto\int_\gamma \frac{F(w)}{w-z}dw\right).

Yes, this looks complicated, but the idea is you plug in a function {F}, and it spits out another function depending on {F} which you obtain by integrating that thing in a loop around the input. You shouldn’t expect this operator to behave in any reasonable way, but if you plug in a holomorphic function {F}, it spits out {2\pi i\cdot F}. That is,

Theorem 2 (Cauchy integral formula) Holomorphic functions are eigenfunctions of the above operator with eigenvalue {2\pi i}.

Thank you to Jordy Greenblatt for pointing this out to me. This result is what makes complex analysis so nice. Remember how eigenvectors were really nice for linear tranformations. This is the basically the same thing.

After that, we used this to prove that holomorphic functions were analytic. I’ll use this fact to prove the following extension:

Theorem 3 (Cauchy integral formula) Let {f\in{\mathcal H}(U)}, and {z_0\in U}, and {\gamma} a loop about {z_0}, as above. Then

\displaystyle f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma\frac{f(w)}{(w-z_0)^{n+1}}dw.

Proof: Expand {f} as a power series about {z_0}. Then

\displaystyle f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots

Taking the {n}th derivative and evaluating at {z_0} yields

\displaystyle f^{(n)}(z_0)=n!a_n

Recalling from our proof that holomorphic functions are analytic,

\displaystyle f^{(n)}(z_0)=n!a_n=\frac{n!}{2\pi i}\int_\gamma\frac{f(w)}{(w-z_0)^{n+1}}dw.



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