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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?

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  • $\begingroup$ 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
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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.

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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 $D$.

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

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