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Show $\not\exists$ $f\in O(\mathbb{C})$ holomorphic such that $f(z)=\overline{z}$ when $|z|=1$. (i.e. There is no curve that agrees with $f(z)=\overline{z}$.)

Proof: Assume that $f(z)=\overline{z}$ is holomorphic when $|z|=1$. Then $f$ must satisfy the Cauchy-Riemann equations (i.e. $u_x=v_y$ and $u_y=-v_x$). Let $f(z)=x-iy$, with $u(x,y)=x$ and $v(x,y)=-y$. Then $$u_x=1\neq-1=v_y \text{ and } u_y=0=-v_x$$ Hence $f$ does not satisfy the Cauchy-Riemann equation, which implies that $f$ is not analytic anywhere.

Now this seems too easy. Is what I did right?

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  • $\begingroup$ What is $O(\mathbb{C})$? $\endgroup$ Commented Feb 1, 2018 at 21:22
  • $\begingroup$ The notation means that $f$ is analytic or holomorphic in $\mathbb{C}$ $\endgroup$ Commented Feb 1, 2018 at 21:24
  • $\begingroup$ "Proof: Assume that $f(z)=\overline{z}$ is holomorphic when $|z|=1$. " That's not what you want to say. $\endgroup$
    – zhw.
    Commented Feb 1, 2018 at 21:30
  • $\begingroup$ How are you computing $u_x$ and $u_y$ if you only know $f$ on the unit circle? As noted below, the function $f(z) = 1/z$ has $f(z) = \overline{z}$ for $|z| = 1$ and is holomorphic on a neighborhood of the unit circle; a local result like the Cauchy-Riemann equations isn't going to be enough. $\endgroup$
    – anomaly
    Commented Feb 1, 2018 at 21:32
  • $\begingroup$ Please use $\not\exists$ only when writing actual formulas. In text, use words. You are communicating with humans. $\endgroup$ Commented Feb 1, 2018 at 22:04

4 Answers 4

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The theorem about Cauchy-Riemann equations that you are using is about functions defined on open sets. It doesn't apply here.

Note that, if $\gamma\colon[0,2\pi]\longrightarrow\mathbb C$ is defined by $\gamma(t)=e^{it}$, then$$\int_\gamma f(z)\,\mathrm dz=\int_0^{2\pi}\overline{e^{it}}ie^{it}\,\mathrm dt=2\pi i.$$If $f$ was analytic then, by Cauchy's theorem, that integral would be equal to $0$.

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  • $\begingroup$ Where did you get this equation? $\endgroup$ Commented Feb 1, 2018 at 21:54
  • $\begingroup$ Do you mean $\int_\gamma f(z)\,\mathrm dz=\int_0^{2\pi}\overline{e^{it}}ie^{it}\,\mathrm dt$? $\endgroup$ Commented Feb 1, 2018 at 22:08
  • $\begingroup$ Yes. That equation $\endgroup$ Commented Feb 1, 2018 at 22:09
  • $\begingroup$ @UsernameUnknown That's the definition of the integral of $f$ along the path $\gamma$. $\endgroup$ Commented Feb 1, 2018 at 22:10
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Just because $f(z) = \bar{z}$ when $|z| = 1$ does not mean that $f(z) = x-iy$ on any open set containing the set where $|z| = 1$. In order to use Cauchy-Riemann equations you need a function defined on an open neighborhood of your point.

I'll also note that $f(z) = 1/z$ does satisfy $f(z) = \bar{z}$ for $|z| = 1$, so any argument needs to use the fact that $f(z)$ is assumed to be holomorphic everywhere inside the disk and not just on the unit circle.

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If you had such a function f, then the product zf would be an entire function whose value on the unit circle is constantly 1. It would follow from this that $zf=1$ on the whole plane and this is obviously impossible.

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Here is another variant:

Suppose $f(z)$ is analytic with $f(z)=\bar{z}$ on the unit circle. Then $g(z) = zf(z)-1$ is analytic, and for $|z|=1$ we have $g(z)=0$. Hence $g(z)=0$ everywhere, but this means $$f(z)=\frac{1}{z}$$
for $z\ne0$, which is impossible.

Note however that this function is analytic in $\mathbb{C}\setminus\{0\}$ and that $f(z)=\bar{z}$ on the unit circle.

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