Assume that a function $f$ is analytic in the disk $\Delta(0,r)$ and that $0<s<r$. Define a path $\gamma:[0,2\pi]\rightarrow \mathbb{C}$ by $\gamma(t)=f(se^{it})$. Demostrate that $$\ell(\gamma)\geq 2\pi s|f'(0)|$$ (Hint. Apply Cauchy's integral fomula, but not to $f$.)
Following the suggestion, I have applied the integral formula of Cauchy to $f'$ and have arrived at the following:
$$f'(z)=\frac{1}{2\pi i}\int_{|w|=s}\frac{f(w)}{w-z}dw$$ so $$f'(0)=\frac{1}{2\pi i}\int_{|w|=s}\frac{f(w)}{w}dw$$ so $$|f'(0)|=\frac{1}{2\pi }\int_{|w|=s}|\frac{f(w)}{w}||dw|\leq \frac{M}{2\pi s}\int_{|w|=s}|dw|$$
But I do not know if what I am doing is correct or if it is what I have to do, I do not know what else to do, could someone help me please? Thank you very much.
Note that by change of variables in the Cauchy formula for $f'$ we have, $$f'(0)=\frac{1}{2\pi i}\int_{|w|=s}\frac{f'(w)}{w}dw \overset{\color{red}{w= se^{it}}}{=}\frac{1}{2\pi i} \int_0^{2\pi} f'(se^{it})|dt$$
On the other hand, $$\gamma(t)=f(se^{it}),\implies\gamma'(t)= sie^{it}f'(se^{it})$$
By definition we know that $$\color{blue}{\ell(\gamma)= \int_0^{2\pi} |\gamma'(t)|dt} = s\int_0^{2\pi} | f'(se^{it})|dt \ge s\left|\int_0^{2\pi} f'(se^{it})|dt \right|=s\left|2\pi if'(0)\right| = \color{blue}{2\pi s\left| f'(0)\right|} $$
We used the first relation above.