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About meromorphic function, wiki says:

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?

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    $\begingroup$ 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. $\endgroup$ – user_of_math Oct 20 '14 at 9:05
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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?).

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  • $\begingroup$ is it because there's a jump at right side of the x axis? $\endgroup$ – athos Oct 20 '14 at 9:19
  • $\begingroup$ Even more: with any possible definition of $\arg$. Do full turn around the origin. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 20 '14 at 9:23
  • $\begingroup$ "Do full turn around the origin" -- i don't get it... $\endgroup$ – athos Oct 20 '14 at 9:25
  • $\begingroup$ Typo correction: "Do a full turn". Like the point $(\cos t,\sin t)$ when $t$ varies from $0$ to $2\pi$. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 20 '14 at 9:28

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