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What's the shortest way to show that there is no analytic function $f$ on $\mathbb{C} \backslash \lbrace 0 \rbrace$ such that $$\exp(f (z)) = z$$ for all nonzero complex numbers $z$?

I came across an old problem set of mine, which answered this. My long argument was basically:

  1. Existence of such a map implies that any open subset of $\mathbb{C} \backslash \lbrace 0 \rbrace$ can be expressed as a disjoint unions of open sets in $\mathbb{C}$, each of which is mapped homeomorphically onto $\mathbb{C} \backslash \lbrace 0 \rbrace$ by the exponential map.
  2. But the preimage of $\mathbb{C} \backslash \lbrace 0 \rbrace$ under the exponential map cannot be such a disjoint union of open sets in $\mathbb{C}$, as described.

Looking back, my answer is too long for my taste as it ran over a page. I think this should have a shorter proof but it escapes me at this time.

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2 Answers

I think the easiest way is the following (which shows that there doesn't even exist a continuous such function on $|z|=1$).

Let $z=e^{i\theta}$. Then $F(e^{i\theta}) = i\theta + 2\pi i k(\theta)$, where $k$ is integer-valued. By continuity, $k$ must be constant, but then $F(e^{0i}) \neq F(e^{2\pi i})$, which is a contradiction.

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Could someone explain this in more detail? –  user58191 Feb 27 '13 at 8:45
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Differentiating the defining equation gives $f'(z)=1/z$ for all $z\ne 0$. Then, integrating $f'(z)$ around the unit circle gives $2\pi i\ne 0$, which is a contradiction.

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This is, by itself, not a contradiction, since the punctured plane is not simply connected. Your $f'$ is analytic on $\mathbb{C}\setminus\{0\}$ despite the fact that it integrates so something non-zero. (It's possible to patch things up of course, but as stated, it looks misleading.) –  mrf Jan 17 '13 at 10:49
    
@mrf, $1/z$ having an antiderivative in a domain would imply that the integral of $1/z$ along any closed curve in the domain to be zero. This has nothing to do with the simply-connectedness of the domain. –  user27126 Jan 17 '13 at 10:51
    
Integrating a derivative over a closed curve can't give a nonzero value, because this contradicts the Fundamental Theorem of Calculus. –  David Moews Jan 17 '13 at 10:52
    
@Sanchez, absolutely. But it would be much cleared if stated. –  mrf Jan 17 '13 at 10:52
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