3
$\begingroup$

There doesnt exist a holomorphic $f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 $, so that $e^{f(z)} = z $ for all $z \in \mathbb C \setminus 0 $

I tried to work with the derivative of f(z) but i didnt manage to get anything out of it.

  • 0
    If there were such an $f$, then we'd have $$\int_{\gamma} f'(z)\,dz = 0$$ for all (piecewise $C^1$) closed curves $\gamma$ in $\mathbb{C}\setminus \{0\}$. What would $f'(z)$ be? – 2017-01-14
  • 0
    $f'(z) = \frac{1}{e^{f(z)}}$ and this function isnt holomorphic – 2017-01-14
  • 0
    Simpler form: $f'(z) = \frac{1}{z}$. That is holomorphic on $\mathbb{C}\setminus \{0\}$. – 2017-01-14
  • 0
    Thank you very much its really late here and i confused myself abit too much. – 2017-01-14

2 Answers 2

2

Assume $f$ exists. By the chain rule, we'd have $$ f'(z)e^{f(z)}=1 $$ so $$ f'(z)=\frac{1}{e^{f(z)}}=\frac{1}{z} $$ Let $\gamma$ be the unit circle: $\gamma(t)=e^{it}$, for $t\in[0,2\pi]$. Then $$ 0=\int_\gamma f'(z)\,dz=\int_\gamma\frac{1}{z}\,dz =\int_0^{2\pi}\frac{1}{e^{it}}ie^{it}\,dt=2\pi i $$ a clear contradiction.

  • 0
    Thank you aswell. I was missing a point in my argumentation – 2017-01-14
2

Here's another proof: Write $f=u+iv.$ Then

$$\ln |z| = \ln |e^{f(z)}| = \ln e^{u(z)} = u(z), \,z\ne 0.$$

Let $\log z$ denote the principal value logarithm in $U=\mathbb C \setminus (-\infty,0].$ Then we have $\text { Re }(f(z) - \log z) \equiv 0$ in $U.$ But an analytic function in $U$ that is purely imaginary is constant. Thus $f(z) - \log z$ is constant in $U,$ which implies $f(z)$ has discontinuities along $(-\infty,0],$ contradicition.