So, $\log z$ is multivalued, but if you remove any ray from the origin to infinity, you get an analytic function. What's it's derivative on any of these branch cuts? Is it always $\frac{1}{z}$?
No, the cut is what you remove. The choice of function (as in this case there are still countably many choices, once you fix the branch cut) on the restricted domain is called a well-defined branch.
@Thorgott calling the two maps to the sphere g_+ and g_-, we want g_+ f = g_- f. but as given g_+ f is the constant map to the north pole and g_- f is the constant map to the south pole
Show that if the analytic function $f(z)$ has a zero of order $N$ at $z_0$, then $f(z) g(z)^N$ for some $g(z)$ analytic near $z_0$ and satisfying $g'(z_0) \neq 0$....By the hypothesis, there exists $h(z)$ analytic at $z_0$ and $h(z_0) \neq 0$. My idea was to take $g(z) = (z-z_0) e^{\frac{\log h(z)}{N}}$. It's clear that $g(z)^N = f(z)$; and if $z_0$, then it's clear that $g'(z_0) \neq 0$...But what if $z_0=0$? How do I deal with this case?
Hey, could you look at this question? It's about maximizing the amount of fish collected when the fish population changes according to logistic growth equations: math.stackexchange.com/q/3874199/595055
my notes say this is an example of a domain in which factorization of non-unit, non-zero elements into irreducibles fails: 'rational polynomial ring' $R$ where elements are finite formal sums $$\sum_{i=1}^k a_i t^{b_i}$$ where $a_i \in \mathbb{C}$ and $b_i \in \mathbb{Q}$, in this ring $t$ is a non-zero non unit with no irreducible factorization, shouldn't this ring be such that $b_i \in \mathbb{Q}_{\geq 0}$ instead? Since otherwise $(t)(t^{-1}) = 1t^0 = 1_{R}$ as far as I can tell
