Why is $\frac{1}{g(z)} \sim \frac{1}{z - z_0} \cdot \frac{1}{g'(z_0)}$ for $z_0$ a pole?

Question

In lecture I encountered the following remark: Let $F(z) := \frac{1}{g(z)}$ with $g(z) := 1-\log \frac{1}{1-z}$. Then $z_0 := 1-e^{-1}$ is a pole and so

$$F(z) \sim \frac{1}{z - z_0} \cdot \frac{1}{g'(z_0)}.$$

I realise that $F$ is a meromorphic function and that the above estimation is similar to the Laurent Series of $F(z)$, but I do not see where the $\frac{1}{g'(z_0)}$ comes from. Could you please explain this to me?

Edit: The Taylor Expansion, as I know it, would look as follows:

$$F(z) = \sum_{n \ge 0} (z-z_0)^n \frac{F^{(n)}(z)}{n!},$$

so I do not understand where the $\frac{1}{z - z_0}$ should come from and additionally $F'(z) \ne g'(z)$.

Answer 1

It's only true if $z_0$ is a simple pole.

Just Taylor expand the denominator $g(z)$ around $z=z_0$ and keep the first nonzero term (which is the one of degree $1$ if $z_0$ is a simple pole).

Answer 2

Another way to look at it: If $g$ has a simple zero at $z_0$ (i.e. $g(z_0) = 0 \ne g'(z_0)$) then $F(z) =1/g(z)$ satisfies $$ (z-z_0)F(z) = \frac{z-z_0}{g(z)-g(z_0)} = \frac{1}{\frac{g(z)-g(z_0)}{z-z_0}} \to \frac{1}{g'(z_0)} $$ for $z \to z_0$, and that means $$ F(z) \sim \frac{1}{z-z_0} \cdot \frac{1}{g'(z_0)} $$ for $z \to z_0$.

Regarding your addendum to the question: $F$ does not have Taylor series at $z_0$ but a Laurent series. In the case of a simple pole that is of the form $$ F(z) = \sum_{n \ge -1} c_n(z-z_0)^n $$ with complex numbers $(c_n)_{n \ge -1}$. But you don't have $c_n = \frac{F^{(n)}(z)}{n!}$. That would make no sense since $F$ and its derivatives all have a pole at $z_0$.