Using Cauchy's Integral Formula to show that $f(z) = e^z$ for every $z$ with $|z| \lt 1$

Question

I'm learning about complex analysis and need some help with this problem :

Given $f : \Bbb C \rightarrow \Bbb C $ analytic with $f(z) = e^z$ for every $z$ with $|z| = 1$. Show that $f(z) = e^z$ for every $z$ with $|z| \lt 1$. (Hint : use Cauchy's Integral Formula for $h(z) = f(z) - e^z$).

Here's my attempt :

Suppose $z \in \Bbb C$ with $|z| \lt 1$. By Cauchy's Integral Formula we have :

$$h(z) = \frac{1}{2\pi i} \int_{|w|=1} \frac {h(w)}{w - z} \, dw \iff f(z) - e^z = \frac{1}{2\pi i} \int_{|w|=1} \frac {f(w) - e^w}{w - z} \, dw$$

but I don't know how to continue from here.

Answer 1

You are done actually. $f(w) - e^w = 0$ for all $w$ in the unit disk, so there it is (you are integrating the zero function). Also, this works for any function, not just the exponential function. The values of a holomorphic $f$ on the boundary of a disk completely determine its values inside the disk.

Even more generally: Identity theorem (see the "An improvement" section).