I feel I incorrectly tackle this problem:
Determine for which values of $p\in \mathbb C$ there is an entire function $f(z)$ satisfying these conditons: $$ |f'(z)| \leq |z| \quad \forall z \in \mathbb C $$ and $f(0)=p$, $~f(1)=0$
My approach: We notice that $f(z)$ must be entire, so we may write: $$ f(z) = \sum_{k=0}^\infty c_n z^n $$ Which converges on all of $\mathbb C$. We know how to derive power series: $$ f'(z) = \sum_{k=1}^\infty n c_n z^{n-1} $$ Which also converges again on all of $\mathbb C$ by an earlier lemma on the derivatives of (complex) power series. We conclude that $f'(z)$ is also entire, as it is analytic , hence holomorphic on all of $\mathbb C$. We observe the condition: $$|f'(z)|\leq 1 \cdot |z^1| $$ implies by the generalised Liouville theorem that $f'(z)$ is a polynomial of at most degree $1$. We thus know for some $a_0, a_1 \in \mathbb C$: $$ f'(z) = a_0 + a_1 z$$ We integrate with respect to $z$ and realise that: $$ f(z) = C+ a_0 z + \frac{1}{2}a_1 z^2 $$ Now observe $f(0)=C=p$. The second condition only gives us that $f(1)=p+ a_0 + \frac{1}{2}a_1=0 \implies p=-a_0 - \frac{1}{2}a_1$
But it feels like I need more restrictions. I suppose it works for $\forall p \in \mathbb C$
I there an error in my reasoning here? I think this answer is rather weird.
If $f'(z)=a_0+a_1z$ and $|f'(z)|\leq |z|$, then in particular at $z=0$ you have $|a_0|\leq |0|$, so $a_0=0$. So, you in fact have $f'(z)=a_1 z$. Again, by the condition $|f'(z)|\leq |z|$, you get $|a_1|\leq 1$. I think those are the missing conditions.