Question: Let $F(z)=z - \frac{p(z)}{p'(z)}$, where $$p(z)= (z-\alpha)^k \prod_{j=1}^{s} (z-\alpha_j)~.$$ Show that $|z-\alpha| < \epsilon$ implies $$|F(z) - \alpha| < |z-\alpha|~.$$ Here $\epsilon > 0$ is arbitrary small.
My approach: $$|F(z) - \alpha| = \left| (z-\alpha) + \frac{p(z)}{p'(z)} \right| \leq |z-\alpha| + \left| \frac{p(z)}{p'(z)} \right|~.$$ Computing the logarithmic derivatives yield $$\frac{p'(z)}{p(z)} = \frac{k}{(z-\alpha)} + \sum_{j=1}^{s} \frac{1}{(z-\alpha_j)}~.$$
I'm sure this is a simple enough exercise but I cannot wrap around my head to get this done. Any help is much appreciated.
I assume you want the conclusion to hold for sufficiently small $\epsilon>0$ (not "arbitrarily small" which does not make sense). The result then holds as long as $p$ is any analytic function in a neighborhood of $\alpha$ which vanishes at $\alpha$.
It suffices to show that $|F'(\alpha)|<1$. We can compute $$F'=1-\frac{p'^2-pp''}{p'^2}=\frac{pp''}{p'^2}.$$ Let us now assume for convenience of notation that $\alpha=0$, and write $p(z)=z^kq(z)$ where $q$ is such that $q(0)\neq 0$ (and $k>0$ by assumption). We then have $$F'(z)=\frac{z^k q(z)(k(k-1)z^{k-2}q(z)+2kz^{k-1}q'(z)+z^kq''(z))}{(kz^{k-1}q(z)+z^kq'(z))^2}=\frac{q(z)(k(k-1)q(z)+2kzq'(z)+z^2q''(z))}{(kq(z)+zq'(z))^2}.$$ Plugging in $z=0$, this reduces to just $$F'(0)=\frac{k(k-1)q(0)^2}{k^2q(0)^2}=\frac{k-1}{k}$$ and so $|F'(0)|<1$ as desired.