$$D=\{z \in \mathbb {C}: |z| \leq 1\}$$

$$S^1 = \{z \in \mathbb{C}: |z|=1\}$$

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?

  • $\begingroup$ The title of your question does not agree with the rest of your question: are you looking at $1/z$, or complex conjugation? $\endgroup$ – Joppy Sep 26 '17 at 0:47
  • $\begingroup$ z^- = 1/z on the circle so it is ok $\endgroup$ – StuartMN Sep 26 '17 at 0:53
  • $\begingroup$ NO. Because then $f(z)=1/z$ for all $z\in D,$ including $z=0,$ which is absurd. $\endgroup$ – DanielWainfleet Sep 26 '17 at 1:13

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.


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 $D$.

  • $\begingroup$ This is fine, since we're assuming that $f$ is analytic on the closed disk, which is to say holomorphic in a neighborhood of the closed disk. But you might note the other two solutions posted work assuming just that $f$ is analytic in the interior and continuous on the closure, a stronger result. $\endgroup$ – David C. Ullrich Oct 24 '17 at 6:04

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.