Use Cauchy's Integral Formula to express $f(0)$ as an integral on the circle $C(0, r)$

Question

Suppose that $f(z)$ is holomorphic on $D(0, 1)$ and let $0 < r < 1$. Use Cauchy’s integral formula to express $f(0)$ as an integral on the circle $C(0, r)$.

I don't get what is being asked here as we know that this function is defined on $C(0, r)$. I thought that the use of Cauchy's Integral Formula was only to get the value of an integral at an undefined point or am I thinking about this wrong and is it just $$f(0) = \frac{1}{2i\pi} \int_{C(0, r)} \frac{f(z)}{z} \,dz$$

Answer 1

The Cauchy integral formula say that:

the value of a function that is analytic inside a closed simple curve $C$ is determined, at any point $z=a$ inside this curve, only by the values of the function on the curve $C$. And more, this is true only for all teh derivatives of the function.

This is is expressed by the formula: $$ f^{(n)}(a)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-a)^{n+1}}dz $$

And this formula is so remarkable just because the values on the boundary determines the values on all internal points!

And, in your case, gives as you noted,

$$f(0) = \frac{1}{2i\pi} \oint_{C(0, r)} \frac{f(z)}{z} \,dz$$