*This seems a very well-known result in complex analysis, still I found it very hard to interpret.
Also f is analytic.
I try to attack this problem in the following way:
We can assume that a is a zero of f of order n, then $f(z) = (z-a)^n g(z)$. Then g : D(a; R) → C is also an analytic function and $g(a) \not= 0.$
But here I am lost, it seems that g may have other zeros in D(a; R), and I don't see how analyticity helps us.
$g$ is continuous. Hence, there exists $\delta >0$ such that $|g(z)-g(a)| <\frac {|g(a)|} 2$ for $|z-a|<\delta$. This implies such that $|g(z)| \geq \frac {|g(a)|} 2$ for $|z-a| <\delta$. There is no zero of $g$ in $B(a,\delta)$.