Why is the algebraic number a whole number. Assume the function $f$, analytic on some domain has a non-essential singularity $a$. Define the algebraic order $h$ of $f$ at $a$ to be the real number such that $\lim_{z\to a}|z-a|^k|f(z)|=0$ for any $k>h$ and $\lim_{z\to a}|z-a|^k|f(z)|=\infty$ for any $k<h$. Why is $h$ a whole number?
 A: A non-essential singularity is either a pole or removable. In either case, we can write
$$f(z) = (z-a)^m\cdot g(z)$$
with a holomorphic function $g$ in a neighbourhood of $a$ with $g(a) \neq 0$ and a uniquely determined $m\in\mathbb{Z}$, unless $f \equiv 0$. You can read that off for example from the Laurent series of $f$.
Then for any $k > -m$ we have
$$\lvert z-a\rvert^{k}\lvert f(z)\rvert = \lvert z-a\rvert^{k+m}\lvert g(z)\rvert,$$
where the first term tends to $0$ since $k+m > 0$, and the second term is bounded (it converges to $\lvert g(a)\rvert$), so the entire expression tends to $0$ for $z\to a$. And for $k < -m$, the second term still converges to a nonzero number, while the first now tends to $+\infty$ since $k+m < 0$.
Either way, the algebraic order of $f$ at $a$ is $-m$. So the reason that the algebraic order is always an integer is analyticity. Since a holomorphic function can be expanded into a power series around each point of its domain, it can only have zeros of integer order (unless it vanishes identically on some component[s] of its domain). And since a pole of $f$ in $a$ becomes a removable singularity of $(z-a)^m\cdot f(z)$ for large enough $m\in\mathbb{N}$, poles also must have integral order.
