# Complex logarithm vs real logarithm

Suppose we are given the function $$\theta = \ln |z| \quad \text{defined on the upper half plane \{ z \in \mathbb{C} \colon \Im( z) > 0\}}$$ Naively, I would go and manipulate $$\ln | z| = \ln (z\bar z)^{\frac{1}{2}} = \frac{1}{2}(\ln z + \ln \bar z + 2\pi i k(z)) \quad \text{(for some integer - valued function k)}$$ and so, because of the last term being integer-valued and non-constant, one cannot take partial derivatives $\partial \theta / \partial z$, $\partial \theta / \partial \bar z$.

On the other hand, the complex logarithm is holomorphic in $\mathbb{C} \setminus (-\infty,0)$ so I ought to be able to differentiate after all since I am only working on the upper half plane.

But one of the problems I have with this reasoning is that $\theta$ is a real - valued function, so if it is holomorphic then it must be constant. There must be an error in my reasoning, what am I doing wrong?

My guess is that I am erroneously mixing up the complex logarithm on the right hand side and the real logarithm on the left hand side, and that $\log \bar z$ is also not the complex logarithm but the composition of the latter with complex conjugation .. Unfortunately I am out of depth with the complex logarithm, my course in Complex Analysis barely scratched the surface of the study of this interesting function.

• It seems to me that you haven't actually chosen a branch of the complex logarithm - of course you can modify a logarithm by $2\pi i k$ and get another logarithm... the idea though is that if you restrict yourself to a specific branch of the logarithm, you get a holomorphic function. – Nick Peterson Jul 13 '13 at 12:03
• $\log z$ is holomorphic, $\log|z|$ isn't. $|z|$ isn't. – Gerry Myerson Jul 13 '13 at 12:09
• Basically, $z\to\bar z$ is not holomorphic. – Thomas Andrews Jul 13 '13 at 12:46
$f\colon z\mapsto \ln \bar z$ is not homolorphic because $f(z+h)=f(z)+\frac1{\bar z}\cdot \bar h + o(h)$ whereas holomorphic needs $f(z+h)=f(z)+a\cdot h+o(h)$.