Why $f’/f$ has a simple pole with residue $k$

Question

I have some confusion in Joseph Bak complex analysis Theorem $10.8$

Theorem $10.8$

Suppose $\gamma $ is a regular closed curve. If $f$ is meromorphic inside and on $\gamma$ and contains no zeroes or oples on $\gamma$ , and if

$\mathbb{Z}$= number of zeroes of $f$ inside $\gamma$ ( a zero of order k being counted $k$ times

$\mathbb{P}=$ number of poles of $f$ inside $\gamma$ (again with multiplicity)

then $$\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathbb{Z}-\mathbb{P}$$

Proof : Note that $f'/f$ is analytic except at the zeroes or poles of $f$ .If $f$ has a zero of order k at z=a ,that is ,if

$f(z)=(z-a)^kg(z)$ with $g(z) \neq 0$

then $f'(z)=(z-a)^{k-1}[kg(z) +(z-a)g'(z)]$ has a zero of order $k-1$ at $z$ and $$\frac{f'(z)}{f(z)}=\frac{k}{z-a}+\frac{g'(z)}{g(z)}$$

Hence , at each zero of $f$ of order $k$ , $f'/f$ has a simple pole with residue $k$

My confusion : why $f'/f$ has a simple pole with residue $k$?

My thinking : At a simple pole $a$ , the residue of $f$ is given by $Res(f,a)=\lim_{z\to a}f(z)$

$$\implies Res(\frac{f'(z)}{f(z)},a)=\lim_{z\to a} (z-a) \frac{k}{z-a}+ \lim_{z\to a} (z-a)\frac{g'(z)}{g(z)}$$
$$\implies Res(\frac{f'(z)}{f(z)},a)=k+ \lim_{z\to a} (z-a)\frac{g'(z)}{g(z)} \neq k$$

Answer 1

$Res(\frac{f'(z)}{f(z)},a)=k+ \lim_{z\to a} (z-a)\frac{g'(z)}{g(z)}$ is actually equal to $k,$ because $g(a) \neq 0,$ so the second summand is $(a-a)\frac{g'(a)}{g(a)} = 0.$