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Is this statement true or false? I see it in a book, but I can not give a counterexample. Could you?

mrf
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user39843
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2 Answers2

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Since $f(z)\ne0$ for $z\in D$ the function $$\ell(z):=\ell_0 +\int_0^z{f'(\zeta)\over f(\zeta)}\ d\zeta$$ is analytic in $D$. The constant $\ell_0\in\Bbb C$ has been chosen such that $e^{\ell_0}=f(0)$. One computes $$\exp\bigl(\ell(0)\bigr)=e^{\ell_0}=f(0)\ ,$$ furthermore $${d\over dz}\bigl(f(z)e^{-\ell(z)}\bigr)=e^{-\ell(z)}\left(f'(z)-f(z){f'(z)\over f(z)}\right)\equiv0\ .$$ It follows that $f(z)=e^{\ell(z)}$ for all $z\in D$, so that one is allowed to call $z\mapsto \ell(z)$ a logarithm of $f$ in $D$. Any function differing from $\ell$ by an additive constant $2k\pi i$, where $k\in\Bbb Z$, shares this logarithmic property with $\ell$.

Christian Blatter
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You need to know that $f$ is holomorphic and that with $\log z$ you mean any logarithm of $z$ i.e. any $w$ such that $e^w =z$. Since the domain of $f$ is simply connected you can uniquely define the logarithm on the graph of $f$.

Emanuele Paolini
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  • Take $f(z)=e^z$. Then $f(\mathbb{C})=\mathbb{C}^*$ is not simply connected... – Julien Mar 03 '13 at 14:13
  • @julien: that's true... I corrected the statement – Emanuele Paolini Mar 03 '13 at 14:19
  • The domain of $f$ contains $0$. And this has nothing to do, anyway, with the existence of a holomorphic $\log f$. – Julien Mar 03 '13 at 14:21
  • There you go. But note that it is not unique, since $\log f+ 2ik\pi$ will do. You might want to add a reference for this powerful fact. Here is one: http://books.google.fr/books?id=FUWPyHM-XK0C&pg=PA127&lpg=PA127&dq=simply+connected+and+log+of+holomorphic+function&source=bl&ots=5nCtxnyFhj&sig=La2ocCYiTKY8AlJiFeWn8iuzhJc&hl=fr&sa=X&ei=qF4zUZ-BIca70AG__oDIAQ&ved=0CEsQ6AEwAw#v=onepage&q=simply%20connected%20and%20log%20of%20holomorphic%20function&f=false – Julien Mar 03 '13 at 14:33