Let $f$ be a holomorphic function on the open unit disk $D$and suppose that $f$ has a zero of order $n$ in $D.$ In a book I am reading it is claimed that, for small $r>0,$ there is a constant $C>0,$ such that $$|f(re^{i \theta})| < C r^n.$$ However, I can not see why this does not hold for any $r>0$ such that $D(0;r) \subset D.$ i.e for $r$ such that the disk of radius $r$ is contained in the unit disk. Is it true that this holds for all such $r,$ or must one in some cases take $r$ to be smaller? If so, how small?
Here are my thoughts. One should look at the power series representation $$f(z) = z^n (\sum_{k=n}^\infty a_n z^{k-n}).$$
Then we have $$|f(re^{i\theta})| \leq r^n \sum_{k=n}^\infty |a_n| r^{k-n}.$$
We thus want to consider $\sum_{k=n}^\infty |a_n|r^{k-n}.$ We have that this infinite sum is bounded by some constant, $C(r)>0.$ Thus,
$$|f(re^{i \theta}| \leq C(r) r^n.$$ For $s < r$ we have $C(s) < C(r),$ so that we can take $C= C(r)$ above and have $$|f(re^{i \theta}| \leq C r^n.$$
I think you are right. I'm not sure that you even need $f$ to vanish to order $n$ in $D$: for any $r>0$ with $D(0,r)\subseteq D$, the function $$\frac{|f(re^{i\theta})|}{r^n} $$ is a continous function of $\theta \in [0, 2\pi]$ hence is bounded, by a bound $C(r)$, depending on $r$.
On the other hand if $C$ is not allowed to depend on $r$, what about $f(z)=z^n/(1-z)$? and consider $r \to 1$.
I was confused in your question whether $C$ can depend on $r$ and whether the inequality holds for a fixed $r$ or for all $r$