Is there an analytical function such that $f (z)=1/z$ for all $z \in S^1$?

Question

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?

Answer 1

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.

Answer 2

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 $f$.

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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 $D$.

Answer 3

You can see this is impossible by applying Maximum Modulus to the function $1-zf(z)$.