This is the statement of the Cauchy Integral Formual. I have seen its proof, and I have used this formula many times to compute integrals. However, I feel like I do not understand this formula completely.
Please note that Wikipedia says, "it expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk." This statement seems very strong and insightful. My question is: What does this exactly mean? I have some sort of vague intuitive understanding but I am not satisfied with my understanding at all. Since the integral of
${\frac {1}{z-a}}$ is $2 \pi i$ along the boundary of the disc, "roughly" we are taking "the average" of the function along the boundary of the disc to compute the function value $f(a)$ for every $a$ in the interior of $D.$ Could you help me understand the following sentence?
"It (the Cauchy Integral Formula) expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk."
Thanks so much.
Best Answer
Note that, when $|a-z_0|<r$, we have$$f(a)=\frac1{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{z-a}\,\mathrm dz.\tag1$$Since$$\int_{|z-z_0|=r}\frac{f(z)}{z-a}\,\mathrm dz=\int_0^{2\pi}\frac{f(re^{it})ire^{it}}{re^{it}-a}\,\mathrm dt,$$which depends only on the values that $f$ takes on the circle centered at $z_0$ with radius $r$, $(1)$ tells us that values that $f$ takes at $a$ (which does not belong to that circle) using only the values that $f$ takes on the circle. In other words, knowing only how $f$ behaves on the boundary of the disk centered at $z_0$ with radius $r$ will allow you to compute the values that $f$ takes inside that disk.