Lef $f\in\mathcal{H}(\mathbb{D}):f(0)=0$ with $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. If there exists $n\in\mathbb{N}$ and $c>0$ such that: $$|f(z)|\leq\frac{c}{(1-|z|)^n},\qquad \text{ for all }z\in\mathbb{D}$$ I want to prove that: $$|f'(z)|\leq \frac{2^{n+1}c}{(1-|z|)^{n+1}}$$ for all $z\in \mathbb{D}$.
I had used the Cauchy's formula for the first derivative: $$f'(z)=\frac{1}{2\pi}\oint_\gamma\frac{f(\zeta)}{(\zeta-z)^2}d\zeta,\qquad\text{ with }\gamma(t)=re^{it}\quad (0\leq t\leq 2\pi) $$ and I get: $$|f'(z)|\leq \frac{c}{2\pi}\oint_\gamma \frac{1}{(1-|\zeta|)^n}\cdot\frac{1}{|\zeta-z|^2}|d\zeta| $$ Anyone hace any ideas for continue? Many thanks
Edit: modified answer taking into account Martin R comment.
Hint:
By Cauchy integral formula "around" $z$, $$ f(z)=\frac{1}{2\pi i}\oint_\gamma\frac{f(\zeta)}{(\zeta-z)^2}d\zeta $$ for any curve $\gamma$ which encloses $z$ exactly once. In particular, we can consider the curve $\gamma$ defined by $\gamma(t)=z+re^{it}$ with $0\leq t\leq 2\pi$. For this curve, we know that for every $\zeta \in \gamma$, $\vert \zeta -z\vert =r$. We also see that $$ \frac{2^{n+1}}{(1-\vert z\vert )^{n+1}}=\frac{1}{\left(\frac{(1-\vert z\vert )}{2}\right)^{n+1}} $$
Using this and Cauchy integral formula around $z$, can you think of a value for $r$ (which could depend on $z$) such that it could help you prove the desired inequality for $\vert f'(z)\vert$? More precisely, given $z$, can you think of a value for $r$ such that for all $\zeta \in \gamma$,
$$ \frac{1}{r(1-\vert \zeta \vert)^n}\leq \frac{1}{\left(\frac{(1-\vert z\vert )}{2}\right)^{n+1}} $$