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Let $f$ be a holomorphic function in the disk $D(0,r)$ of center $0$ and radii $r>1$ s.t. for all $z$ of module $1$ we have $|f(z)|<1$.

Show that there is an $a\in D(0,1)$ s.t. $f(a)=a$ and $f'(a) \neq 1$.

Well, so far I've shown that there is a unique $a\in D(0,1)$ s.t. $f(a)=a$.

Indeed, since $|z-(f(z)-z)|=|f(z)| < 1 = |z|$ for all $z\in \partial D(0,1)$ Rouche's theorem says that $f(z)-z$ has the same number of zeros as $z$ in $D(0,1)$, that is one.

Now, a hint on how to prove that $f'(a) \neq 1$?

I was thinking that Schwarz's lemma has similar hypothesis and a pertinent conclusion, but I don't see how to transform $f$ in a holomorphic function with a zero in $0$ still contained in the unit disk and with the same derivative as $f$.

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    If you want to use the Schwarz lemma, look at $T_a \circ f \circ T_a^{-1}$, where $T_a(z) = \dfrac{z-a}{1 - \overline{a}z}$. – 2017-01-22

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If a=0, then you can use the Schwarz lemma. If $a\neq0$, you can use the Blaschke factor $$b_a=\frac{z-a}{1-\bar{a}z}$$ defined on $D(0,1)$. $b_a(0)=a$ and $b_a(a)=0$. Now consider $g=b_a ∘ f ∘ b_a^{-1}$ defined on $D(0,1)$. Then $g(0)=0$. Apply Schwarz Lemma to $g$. Then $|g'(0)|\leqslant 1$, and $$g'(0)=\frac{f'(a)}{\left(1-|a|^2\right)^2},$$ which gives the result as $a\neq0$.

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    Due to chain rule, $$g'(0) =b_a'(0)\cdot f{'}\circ b_a(0) \cdot b_a^{-1}{'}\circ f\circ b_a (0) = b_a{'}(0) \cdot f{'}(a) \cdot b_a^{-1}{'}(a) = \frac{b_a'(0)}{b_a{'}(0)}\cdot f{'}(a)=f'(a)$$, amirite? – 2017-01-22
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    I am sorry, will just make the correction. – 2017-01-23
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A solution without using Schwarz's lemma:

We already knew that $f(z)-z$ had a single zero in $D(0,1)$. Now, that implies that $(f(z)-z)'_{z=a}=f'(a)-1 \neq 0$, or otherwise the zero would have multiplicity of at least two.

Thus $f'(a)\neq 0$.