Two answer the question if a Logarithm of a holomorphic function exists I know the following creterions:

a) If $G\subset \mathbb{C}$ is a simple connected domain and $f$ holomorphic with $f(z)\ne 0$ in $G$. Then it exists a Logarithm of $f$ in $G$.

b) If for the holomorphic function on $G \subset \mathbb{C}$ it is $f(G)\subset \mathbb{C}\setminus \mathbb{R}_{\leq 0}$, then $\log(f(z))$ is a Logarithm of $f$.

Now I have problems with the follow exercise:

Let $\Omega = \mathbb{C}\setminus[-1,1]$. Then there are two functions given:

  1. Show, that on $\Omega$ there does not exist a holomorphic Logarithm of $f(z)=\frac{1}{z^2-1}$.

  2. Show that on $\Omega$ there exists a holomorphic Logarithm of $h(z)=i\frac{z+1}{z-1}$.

But as $\Omega$ is not simple connected domain, I cannot use a). And b) in general is not very practible, I think.

Can anyone please help me? Thank you! :)

  • $\begingroup$ $\log f$ is holomorphic on $U$ iff $\frac{f'}{f}= (\log f)'$ is holomorphic on $U$ and $\log f(s) = \log f(a)+\int_a^s \frac{f'(z)}{f(z)}dz $ doesn't depend on the path $a \to s$ iff for all closed curve $\int_\gamma \frac{f'(z)}{f(z)}dz = 0$ $\endgroup$ – reuns Oct 29 '17 at 14:38
  • $\begingroup$ Yes, true. But $f'(z)=-\frac{1}{(z^2-1)^2}$, therefore $\frac{f'}{f}=-\frac{1}{z^2-1}$. Every closed curve in $\Omega$ incloses the two singularities $z=\pm 1$ with residues $Res(\frac{f'}{f},-1)=-0.5$, $Res(\frac{f'}{f},1)=0.5$. By the Residue theorem it would follow that the integral is zero for every curve, and it would exist a Logarithm. Where is my mistake? $\endgroup$ – Lympstone95 Oct 29 '17 at 14:44
  • $\begingroup$ My $f'$ is wrong. Okay, thanks! $\endgroup$ – Lympstone95 Oct 29 '17 at 14:49
  • $\begingroup$ To compute $f'/f$ factorize $f$ and differentiate $\log f$ $\endgroup$ – reuns Oct 29 '17 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.