Series expansion of $\ln z$ Is there a series expansion of $\ln z$ at $z = 0$, as in $ \ln z = \sum_{n=-\infty}^{\infty} a_{n} z^{n}$? If so, $a_{n} = ?$
It has to be a Laurent series because $\ln z$ diverges at $z = 0$.
 A: $\log(z)$ cannot be continuously well-defined in a neighborhood of $z=0$.
The imaginary part of the logarithm of a point on any circle increases by $i$ times the clockwise angle (in radians) traversed.  Thus, if you circle the origin once counter-clockwise, the logarithm increases by $2\pi i$. To get back to where it was before at that point, we need to discontinuously subtract $2\pi i$ from the logarithm.
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This is why there needs to be a branch cut starting at $0$ when defining $\log(z)$.
So the answer to your question is no, there is no Laurent series for $\log(z)$ at $z=0$.
A: Not only do you have the issue with the singularity at $z = 0$, but because the function is multi-valued, you have to make a branch cut somewhere, and the function won't be continuous in any annulus centered at $z = 0$.
You can write a series centered at, e.g., $z = 1$, or you can write a series for $ln(z - 1)$ centered at $z=0$, but the answer to your original question is 'no'.
A: The singularity of $\ln$ at $z=0$ is not a pole, so the answer is no.
