find $\int _\gamma \frac{f'(z)}{f(z)}dz$ 
let $f$ be an holomorphic function such that $a$ is a zero of order $2$ on $D_f$. find $\int _\gamma \frac{f'(z)}{f(z)}dz$ where $\gamma$ is a circle centered in $a$ contained in $D_f$ and such that the other zeroes of $f$ all lie outside of the circle limited by $\gamma$.

I know that since $a$ is a zero of order $2$ of $f(z)$, then it is a zero of order $1$ of $f'(z)$. therefore function $f'(z)/f(z)$ has a pole of order $1$ at $a$.
I suspect I need to use the residue theorem, but how can I find the residue of $f'(z)/f(z)$?
 A: By the Argument Principle (http://en.wikipedia.org/wiki/Argument_principle)
$$\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}\,dz=N-P=2-0=2,$$
where $N$ and $P$ denote the numbers of zeros and poles, respectively, with each zero and pole counted as many times as its multiplicity and order. 
A: Hint: Write
$$
f(z)=(z-a)^2g(z)
$$
where $g(a)\ne0$. Then
$$
\begin{align}
\frac{f'(z)}{f(z)}
&=\frac{2(z-a)g(z)+(z-a)^2g'(z)}{(z-a)^2g(z)}\\
&=\frac2{z-a}+\frac{g'(z)}{g(z)}
\end{align}
$$
A: Hint: $f(z)=(z-c)^kg(z)$ with $g(c)\ne 0$ when $f$ has a zero of multiplicity $k$ in $c$.
A: Lets generalize. Let $f$ be a holomorphic function such that $a$ is the only zero interior the closed curve $\gamma$. Let $a$ be a zero of order $k$. Then since $f$ is holomorphic there exists $g(z)$ analytic and such that $f(z)=(z-a)^{k}g(z)$ where $g(a)\neq 0$. Therefore $f'(z)=k(z-a)^{k-1}g(z)+(z-a)^{k}g(z)$. Now can you compute 
$$\int_{\gamma}\dfrac{f'(z)}{f(z)}dz=\int_{\gamma}\dfrac{k(z-a)^{k-1}g(z)+(z-a)^{k}g(z)}{(z-a)^{k}g(z)}dz?$$
