Is there an analytical function such that $f (z)=1/z$ for all $z \in S^1$? Let:
$$D=\{z \in \mathbb {C}: |z| \leq 1\}$$
$$S^1 = \{z \in \mathbb{C}: |z|=1\}$$
Im wondering if it is possible to an analytical function $f:D \rightarrow \mathbb{C} $ satisfy $f(z)=1/z \ \ \forall z \in S^1 $.
I guess it is not possible. I tried to use the Maximum\Minimum Modulus Principle, but all I conclude is that $f$ must have zeros in $D$.
Any hints?
 A: If $f: D \to \mathbb{C}$ is analytic, then by Cauchy's integral theorem the contour integral $\oint f \, dz$ taken along $S^1$ vanishes. But the values of $f$ along this curve agree with $1/z$, and we know $\oint 1/z \, dz = 2 \pi i$. So $f$ cannot be analytic.
A: Let $P \in S^1$. Every neighborhood of $P$ has infinitely many roots of the function $f(z) - \frac{1}{z}$, so by looking at the Taylor series about $P$ you can prove that $f(z) - \frac{1}{z}$ is the zero function near $P$.
Since $D \setminus \{ 0 \} $ is connected, by analytic continuation, $f(z) - \frac{1}{z} = 0$ everywhere on $D \setminus \{ 0 \}$. And consequently, $f$ must have a simple pole at $0$, contradicting the hypothesis that $0$ is in the domain of $f$.

More generally, given:


*

*A connected domain $D$

*A sequence of points $s_n \in D$ that converge to a point in $D$

*Two analytic functions on $D$ that agree on every $s_n$


then the two given functions are the same everywhere in $D$.
A: You can see this is impossible by applying Maximum Modulus to the function $1-zf(z)$.
