Show that for all $n ∈ \mathbb N$ there exists a constant $C_n > 0$ such that $|f ^{(n)} (z)| ≤ C_n/(Imz)^ n $ for all $z ∈ H$

Question

Let $H$ be the upper half plane and let $f : H → \mathbb C$ be holomorphic on $H$. Suppose that $f$ is bounded. Show that for all $n ∈ \mathbb N$ there exists a constant $C_n > 0$ such that $|f^{(n)}(z)| ≤ C_n/(Imz)^n$ for all $z ∈ H$

My attempt:

By Cauchy Estimates, we know that $|f^{(n)}(z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|} |f(z)|$ $(*)$

But $f$ is bounded on $H$, then there is $M >0$ s.t $|f(z)|≤M$ for all $z \in H$.

Then $(*)$ becomes:

$|f^{(n)}(z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|} M = \frac{n!}{r^n} M_r$

But I don't know how to continue from here.

Any answers please?

Answer 1

Note that, since $z\in H$, $\text{Im}\ z=|\text{Im}\ z|,$ and of course $|\text{Im}\ z|\leq r.$ Then, $$\frac{1}{r^n}\leq\frac{1}{(\text{Im}\ z)^n}.$$