Why isn't the complex logarithm $\log z$ holomorphic on $\mathbb C -\{0\}$?

Question

Why isn't the complex logarithm $\log z$ holomorphic on $\mathbb C -\{0\}$? Why can't you just say take the $\arg z$ to be in $[0,2\pi)$ and then you don't have to worry about it being a multivalued function.

Answer 1

It's not even continuous. Compare $\log(1-0.00001i)$ and $\log(1+0.00001i)$.

Answer 2

Above axis $\mathbb{R}_{>0}\times\{0\}$, the argument you defined will be close to $0$, whereas under it, it will be close to $2\pi$, so the logarithm you defined that way is not even continuous.

Answer 3

Hint: Find $$\lim_{z \to 1 \\ \arg(z)>0} \log(z) \\ \lim_{z \to 1\\ \arg(z)<0} \log(z) $$

Answer 4

The logarithm of a complex number depends on the arg function. If you start following a circle around the origin starting at a real number $r$, the arg function starts growing from zero until it nears $2\pi$ when it is finishing a full turn. In consequence, the arg function cannot be continuous on any circle that surrounds the origin, and neither can the logarithm.

Answer 5

The mean old exponential function is periodic with period $2\pi i$. The result: much of complex analysis is the study of the complex log function.