1) Suppose I want to find the order of the zero of the following function $$ f(z) = (e^z - 1)^3$$ at $z=0$.
I first find the Taylor expansion for $e^z - 1$, and then write $$e^z - 1 = zg(z),$$ where $g$ is analytic and $g(0) \ne 0$. Next, I say $f(z) = z^3 [g(z)]^3$, so it has a zero of order 3 at 0. Is that right? The reason I'm asking this is because in complex analysis, it is not true that $(z_1 z_2)^\alpha = z_1^\alpha z_2^\alpha$. If it's not right, is there any way I can do it fast without expanding everything?
2) How can I find the order of the zero of $f(z) = [\log(1 + \sin z)]^2$ at $z = 2\pi$ without applying the formula to the whole function $f$ to find the Taylor series? Will the multi-valued $\log$ function affect how I do it?
Thank you.
If we denote the $n$-th derivate of $f(z)$ with $f^n(z)$, then the following holds :
If $f(z)=0$ and $n$ is the smallest positive integer with $f^n(z)\ne 0$ , then the zero $z$ of $f(z)$ has order $n$.