# Complex Calculus - Cauchy integral Formule

I could not prove below theorem. can you help me ?

Thm : let $f(z)$ be analytic in $|z|<1$ with $f(0)=0$ and $\left|f(z)\right|\le 1$ for all $z$, $|z|<1$. prove that $\left|f''(0)\right|\le2$

Thanks

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We apply cauchy's integral formula: we have for $r\in(0,1)$ $$f''(0)=\frac 1{2\pi i}2\int_{C(0,r)}\frac{f(\xi)}{\xi^3}d\xi$$ hence $$|f''(0)|\leq \frac 1{\pi}\frac{2\pi r}{r^3}=\frac 2{r^2}$$ and you get the result taking the limit $r\to 1$.
With $f(z)=z^2$, we have equality.