Problem : If $f(z)$ has an isolated singularity at $z=z_0$ and if $\lim_{z\to z_0} (z-z_0)^{\alpha}f(z)=M\neq0,\infty$, then prove that $\alpha$ must be an integer.
This is Exercise 9.22.1 in Silverman's "Complex Variables with Applications".
Most of the exercises in this book are quite easy, but I have no idea with this question. How should I attempt?
P.S. There is no assumptions about $\alpha$, but I think it should be any complex constant.
P.P.S. When $\alpha$ is not an integer, even though I choose the principal branch of $(z-z_0)^{\alpha}$, does $\lim_{z\to z_0} (z-z_0)^{\alpha}$ exist?
Best Answer
Suppose $a>0$. The hypothesis implies that $|f(z)| \to \infty $ as $z \to z_0$. This property is a characterization of poles. Hence $f$ has a pole at $z_0$. If the order of the pole is $n$ then $f(z) (z-z_0)^{n} $ tends to a finite non-zero limit. Can you now see that $\alpha =n$?.
If $a<0$ use a similar argument. Here $f$ has a zero of some finite order at $z_0$ and the argument is similar.