I'm stuck with this exercise of my "Cauchy-Riemann equation" problem list.

Let $\Omega=\{x\in\mathbb{C}\ | 0<Re(z)<1\}$. Find a determination of the logarithm such that $f(z)=log(z)\cdot log(1-z)$ is holomorphic over $\Omega$.

I know that I have to check for what values of $\theta=arg(z)$ it holds that $$\frac{\partial f}{\partial \bar{z}}=0.$$

I've tried it but I'm not able to simplify the monster after finding $\frac{\partial f}{\partial \bar{z}}$, and I think that I'm missing something.

Thank you.


Hint : multiplication of holomorphic functions is holomorphic. So you need to find a determination of $\log(z)$ which is holomorphic on $ \{z \in \Bbb C : \Re (z) > 0\}$ (for example, it's enough to find such determination on $\Bbb C \backslash \Bbb R_-$).

Do similarly with $\log(z-1)$ and multiply them together.

  • $\begingroup$ But then, $\frac{\partial log(z)}{\partial \bar{z}}=0$ for every $\theta$ such that $z\in\Omega$, and exactly the same for $\frac{\partial log(1-z)}{\partial \bar{z}}$ $\endgroup$ – Relure Jun 18 '17 at 15:38
  • $\begingroup$ Exactly so the product of log will be holomorphic $\endgroup$ – user171326 Jun 18 '17 at 15:40
  • $\begingroup$ as $\partial_{\overline z} fg = (\partial_{\overline z} f) \cdot g + f \cdot (\partial_{\overline z} g ) = 0$ if $\partial_{\overline z} f = \partial_{\overline z} g = 0$. $\endgroup$ – user171326 Jun 18 '17 at 15:41
  • $\begingroup$ So I can take infinite determinations where $f(z)$ is holomorphic right? What's the point behind this exercise then? $\endgroup$ – Relure Jun 18 '17 at 15:42
  • $\begingroup$ Yes exactly. I don't really know what's the purpose of the exercise ... $\endgroup$ – user171326 Jun 18 '17 at 15:43

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