- Assume $f$ is an entire function and for every $z$,$\left|f\left(z\right)\right|>1$,prove that $f$ is a constant function.
- Assume $f$ is analytic on $\left|z\right|<1$ and for every $z$ on $\left|z\right|<1$,$\left|f\left(z\right)\right|\le1$,Prove that $\left|f'\left(0\right)\right|\le1$.
- Assume $f$ is a entire function and $A$ is a fixed number and for every $z$,$\left|f\left(z\right)\right|<A\left|z\right|$,prove that for every $z$, either $f(z)=0$ or $f(z)=az$ ($a \ne 0$)
For the first one I tried the function $e^z$, Since it's entire so the function composition $e^{f(z)}$ is also entire, on the other hand $$\left|e^{f\left(z\right)}\right|\le e^{\left|f\left(z\right)\right|}$$
But I cannot find a bound so I cannot use Liouville's theorem, So there should be a function composition which enables us to use the fact that $\left|f\left(z\right)\right|>1$.
For the second one, I know that if $f$ is analytic on $\left|z-z_{0}\right|\le r$ and is bounded by $M$ on the boundary $\left|z-z_{0}\right|= r$, then $$\left|f^{\left(n\right)}\left(z_{0}\right)\right|\le\frac{n!}{r^{n}}M$$
However $f$ is only analytic on $\left|z\right|<1$ and not on the boundary $\left|z\right|=1$, So I cannot use Cauchy's inequality.
For the third one, since $f$ is entire, so it's analytic on $\left|z-z_{0}\right|\le r$ ( for arbitrary $z_0 \in \mathbb C$), hence by Cauchy integration formula $$f\left(z_{0}\right)=\frac{1}{2\pi i}\int_{\left|z-z_{0}\right|=r}^{ }\frac{f\left(z\right)}{z-z_{0}}dz$$
So $$\left|f\left(z_{0}\right)\right|=\frac{1}{2\pi}2\pi r\frac{\left|f\left(z\right)\right|}{r}<A\left|z\right|\le A\left(\left|z-z_{0}\right|+\left|z_{0}\right|\right)=A\left(r+\left|z_{0}\right|\right)$$
Now I tried to show that either $f(z_0)=0$ or $f(z_0)=az$ for nonzero $a$, But I cannot proceed.
I'm not necessarily looking for the full answer, Rather a hint in the comment section would be appreciated.
For 1) apply Liouville's Theorem to $\frac 1 f$.
For 2) apply the inequality you have written to the disc of radius $r$ around $0$ where $0<r<1$. In the result you get let $r $ increase to $1$.
In 3) strict inequality is wrong since you cannot have $|f(0)|<0$. It must read $|f(z)| \leq A|z|$. Note that $f(0)=0$. Now let $g(z)=\frac {f(z)} z$ for $z \neq 0$ and $g(0)=f'(0)$. Show that $g$ is entire and apply Liouville's Theorem.