Show $\not\exists$ $f\in O(\mathbb{C})$ holomorphic such that $f(z)=\overline{z}$ when $|z|=1$.

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?

Solution 1:

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 $0$.

Solution 2:

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.

Solution 3:

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.