Applying analytic continuation

Question

I'm reading the Stein-Shakarchi. In Chapter 3 there is a problem and I need to check my solution:

Show that there is no function $f$, such that is holomorphic, defined on the open unit disk that extends continuously to its border, and is equal to the function $g(z)=\frac{1}{z}$ in the border.

Here is my "solution": Extend the domain of $f$ in such a way that now is $\mathbb{C}$. Now is $\frac{1}{z}$ if $|z|\geq1$, and is equal to $f(z)$ otherwise. By Morera's theorem every triangle has integral egual to 0 at the border of $\mathbb{D}$, so is holomorphic, in the other hand, f and the new g agrees on a set with accumulation point, so they must be equal. Finally, $g$ is not holomorphic in all $\mathbb{C}$, and $f$ is, so we get a contradiction. Is something wrong here? Do you have any other solution?

Answer 1

If $f\in\mathcal H(\mathbb D)\cap C\big(\overline{\mathbb D}\big)$, where $\mathbb D$ is the open unit disk in $\mathbb C$, then for every $r\in (0,1)$ $$ \int_{|z|=r} f(z)\,dz=0\quad\text{and hence}\quad \int_{|z|=1} f(z)\,dz=\lim_{r\nearrow 1}\int_{|z|=r} f(z)\,dz=0. $$ But $$ \int_{|z|=r} \frac{dz}{z}\ne 0. $$

Answer 2

One can, for instance, extend the well-known identity \[ \sin^{2}(z) + \cos^{2}(z) = 1, \] which is valid originally for z on the real line, to the whole complex plane by analytic continuation.

And analytic continuation: If f is analytic in U, g is analytic in V, and if there is an intersection of U and V, call it S, which contains a segment, on which f = g, then f can be extended to the union of U and V.

The unit circle is a segment.