This is from Stein and Shakarchi.
Give another proof of the Cauchy integral formula
$f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta-z}d\zeta$
using homotopy of curves.
(Hint: Deform the circle $C$ to a small circle centered at $z$, and
note that the quotient $\frac{f(\zeta)-f(z)}{\zeta-z}$ is bounded)
I found other posts about this problem but I couldn't find the answer to my question.
I found that the problem reduces to the problem of showing that $\frac{1}{2\pi i}\int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta$ approaches to zero as $r$ goes to zero. I think this is where I need to use the theorem of homotopy in the book (integrals over holomorphic function on two path-homotopic curves is same). But since end points of different circles are not exactly the same, they are not path-homotopic. How can I modify the theorem to apply it in the problem?
Best Answer
Let $C$ be circle centered at $z_0$ and let $C_r$ be a circle centered at $z_0$ contained within $C$. We consider
$$\frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta-z_0}d\zeta = \frac{1}{2\pi i} \int_{C} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta + \frac{f(z_0)}{2\pi i} \int_{C} \frac{1}{\zeta-z_0}d\zeta$$
Now, observe that $C$ can be continously deformed into $C_r$. Thus, we have
$$\frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta-z_0}d\zeta=\frac{1}{2\pi i} \int_{C_r} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta + \frac{f(z_0)}{2\pi i} \int_{C_r} \frac{1}{\zeta-z_0}d\zeta$$
Now, we prove that the first integral approaches zero as $r \rightarrow 0$:
$$\bigg| \int_{C_r} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta \bigg| \leq \int_{C_r} \bigg|\frac{f(\zeta)-f(z_0)}{\zeta-z_0} \bigg| d \zeta \leq M \ell(C_r)=2\pi Mr \rightarrow 0$$
when $r \rightarrow 0$. The last integral is equal to $f(z_0)$, which proves the theorem.