Let $f$ be a complex polynomial of degree $d \geq 2$ with a simple root at $0$. Assume further that its other roots $z_{1},\cdots,z_{d-1}$ satisfy $\min_{1\leq i\leq d-1}(|z_{1}|,\cdots,|z_{d-1}|)=1$, and that $|f'(0)|<1/8$. I want to prove that there exists a non-zero critical value of $f$ on the open disk $D(0,1/2)$.
I know how to prove that there is a critical value on the disk, possibly equal to $0$. If we assume otherwise, then we can define an inverse $f^{-1}:D(0,1/2)\rightarrow U$, where $U$ is the connected component of $f^{-1}(D(0,1/2))$ containing $0$. But then $U$ does not contain $\overline{D}(0,1)$ since $f$ has a root on the unit circle, and so by Koebe's $1/4$ theorem $|(f^{-1})'(0)|\leq 8$, a contradiction.
There are polynomials satisfying the hypotheses such that $U$ contains a double root of $f$, like $f(z)=z(z-1)^{2}/9$, but they also have another critical point on $U$.
How does one go about proving that $0$ cannot be the only critical value? In case the result I want to prove is wrong, is there a constant that can replace $1/8$ (possibly depending on the degree) that makes it correct?
Edit [incorrect proof]: I have a proof with the extra assumption $|f'(0)|<1/2^{d}$. In "Dynamics in one complex variable", Lemma 8.5, Milnor proves that the largest disk on which $f$ is linearizable has a critical point in its boundary. This disk is clearly contained in $D(0,1)$, and our extra assumption ensures that $f(\overline{D(0,1)})\subset D(0,1/2)$ (write $f(z)=f'(0)z(1-z/z_{1})\cdots(1-z/z_{d-1})$).
I'm still interested in the original case, without the extra assumption. Perhaps a counterexample where $f$ is a rational function (without poles on $D(0,1)$) would be helpful. Indeed, Lemma 8.5 would still apply, but I imagine it could take many iterations before the critical point reaches $D(0,1/2)$.
Edit 2: I made a mistake in my previous edit. Milnor actually proves the following: there is a holomorphic map $\phi$ defined on the basin of attraction of $0$ such that $\phi\circ f=f'(0)\phi$, which is biholomorphic near $0$. If $D(0,r)$ is the largest disk on which $\phi^{-1}$ is defined,then $f$ has a critical point $\zeta$ on the boundary of $\phi^{-1}(D(0,r))$, which is attracted to $0$. Of course, there is no reason to believe that $|\zeta|\leq1$. Milnor's result does imply that some iterate $f^{\circ n}(\zeta)$ is a non-zero element of $D(0,1/2)$, but I'm not sure we can assume that $n=1$.
Here's my second attempt at proving the result. The polynomial $f$ has roots on the unit circle and at the origin, so by the Grace-Heawood theorem $f$ has a critical point $\zeta$ such that $$|\zeta|\leq \frac{1+\cot(\pi/d)}{2}$$ And a sufficiently small $f'(0)$ would send $\zeta$ to $D(0,1/2)$. However, I don't know if a can assume that $f(\zeta)\neq0$.