Question about poles and zeros of a general $f(z)$ I stumbled across this :

Proof of Hadamard - de la Vallée Poussin's Theorem: First recall a basic fact: $$\lim\limits_{z\rightarrow a} \frac{(z-a)f'(z)}{f(z)}=\left\{\begin{matrix} -m, & \text{if } z=a \text{ is a pole of order } m\geq1 \text{ of } f(z)\\  m & \text{if } z=a \text{ is a zero of multiplicity } m\geq1 \text{ of } f(z)\\  0 & \text{if } f(z) \text{ is analytic at } z=a \text{ and } f(a)\ne0 \end{matrix}\right.$$

they say it's a basic fact, but I am not sure where it comes from. Thanks for any justification. 
 A: *

*If $z=a$ is a pole of order $m\geq1$ of $f(z)$ then, from the definition
$$f(z)=\sum\limits_{k=-m}^{\infty}a_k(z-a)^k, a_{-m}\ne0$$
Then 
$$\color{red}{\lim\limits_{z\rightarrow a}\frac{(z-a)f'(z)}{f(z)}}=
\lim\limits_{z\rightarrow a}\frac{(z-a)\sum\limits_{k=-m}^{\infty}ka_k(z-a)^{k-1}}{\sum\limits_{k=-m}^{\infty}a_k(z-a)^k}=\\
\lim\limits_{z\rightarrow a}\frac{\sum\limits_{k=-m}^{\infty}ka_k(z-a)^{k}}{\sum\limits_{k=-m}^{\infty}a_k(z-a)^k}=
\lim\limits_{z\rightarrow a}\frac{(z-a)^m\left(\sum\limits_{k=-m}^{\infty}ka_k(z-a)^{k}\right)}{(z-a)^m\left(\sum\limits_{k=-m}^{\infty}a_k(z-a)^k\right)}=\\
\lim\limits_{z\rightarrow a}\frac{-ma_{-m}+\sum\limits_{k=-m+1}^{\infty}ka_k(z-a)^{k+m}}{a_{-m}+\sum\limits_{k=-m+1}^{\infty}a_k(z-a)^{k+m}}=\frac{-ma_{-m}}{a_{-m}}=\color{red}{-m}$$





*If $f(z)$ is analytic at $z=a$ and $f(a)\ne0$ then it has a Taylor series representation in a neighbourhood of $a$ 
$$f(z)=\sum\limits_{k=0}^{\infty}a_k(z-a)^k, f(a)=a_0\ne0$$
Then
$$\color{red}{\lim\limits_{z\rightarrow a}\frac{(z-a)f'(z)}{f(z)}}=
\lim\limits_{z\rightarrow a}\frac{(z-a)\sum\limits_{k=1}^{\infty}ka_k(z-a)^{k-1}}{\sum\limits_{k=0}^{\infty}a_k(z-a)^k}=\\
\lim\limits_{z\rightarrow a}\frac{(z-a)a_1+\sum\limits_{k=2}^{\infty}ka_k(z-a)^{k}}{a_0+\sum\limits_{k=1}^{\infty}a_k(z-a)^k}=\frac{0}{a_0}=\color{red}{0}$$





*If $z=a$ is a zero of multiplicity $m\geq1$ of $f(z)$ - left as an exercise
A: Hint:  $\frac{f'}{f}$ suggests logarithmic differentiation.  Now what happens if $f(z) = (z-a)^m g(z)$, where $g$ is holomorphic at $a$ and $m$ is any integer?
