# Why the complex logarithm function$\ln(z)$ is not meromorphic on the whole complex plane

In the mathematical field of complex analysis, a meromorphic function on an open subset D of the complex plane is a function that is holomorphic on all D except a set of isolated points (the poles of the function), at each of which the function must have a Laurent series.

, and as an example, it says

The complex logarithm function $f(z) = \ln(z)$ is not meromorphic on the whole complex plane, as it cannot be defined on the whole complex plane while only excluding an isolated set of points.

I don't understand the idea here, why $\ln(z)$ is not meromorphic.

Is this because the complex exponential function is not injective, hence $\ln(z)$ has many many branches? If so, if we limit to the principal value $\text{Log } z$ where the logarithm imaginary part lies in the interval $(−\pi, \pi]$, would that $\text{Log } z$ be meromorphic?

• Meromorphic functions can only have isolated pole singularities. A branch cut is a line, not an isolated singularity. Ergo, $\ln(z)$ cannot be a meromorphic function. – user_of_math Oct 20 '14 at 9:05

Take $z=re^{i\theta}$. If we want a $\log$ function with the expected properties, $$\log z=\log r+i\theta=|z|+i\arg z.$$ But isn't possible define an $\arg$ function without a jump (why?).
• Even more: with any possible definition of $\arg$. Do full turn around the origin. – Martín-Blas Pérez Pinilla Oct 20 '14 at 9:23
• Typo correction: "Do a full turn". Like the point $(\cos t,\sin t)$ when $t$ varies from $0$ to $2\pi$. – Martín-Blas Pérez Pinilla Oct 20 '14 at 9:28