Let $G\subset\mathbb{C}$ a domain and $D=D_1(0)\subset\subset G$. Let $f: G\to\mathbb{C}$ holomorphic and has two roots $z_1, z_2$ with order of zero $k_1, k_2 >0$. For $z\neq z_1, z_2$ let $f(z)\neq 0$. Prove this equation: $$\frac{1}{2\pi i}\int_{\partial D} \frac{f'(\zeta)}{f(\zeta)}d\zeta = k_1+k_2$$
I'm really stucked with this task. I can see a similarity to the Cauchy Integral formula. And I know that a antiderivative of this function is $\log(f(\zeta))$. But I don't know how to get a good proof. Any hints or help? Thank you!
Since $z_1, z_2$ are the only zeros of $f$, $$ f(z) = (z-z_1)^{k_1} (z-z_2)^{k_2} g(z) $$ for some holomorphic function $g: G \to \Bbb C$ which has no zeros in $G$. It follows that $$ \frac{f'(z)}{f(z)} = \frac{k_1}{z - z_1} + \frac{k_2}{z - z_2} + \frac{g'(z)}{g(z)} $$ and therefore $$ \frac{1}{2\pi i}\int_{\partial D} \frac{f'(\zeta)}{f(\zeta)}d\zeta = \frac{k_1}{2\pi i}\int_{\partial D} \frac{d\zeta}{\zeta - z_1} + \frac{k_2}{2\pi i}\int_{\partial D} \frac{d\zeta}{\zeta - z_2} + \frac{1}{2\pi i}\int_{\partial D} \frac{g'(\zeta)}{g(\zeta)}d\zeta \, . $$ The last integral is zero because $g'/g$ is holomorphic in $D$, and the other two can be computed easily.