This seems $\textbf{somewhat related}$ to a Miklos Schweitzer $(1972)$ problem.
$\textbf{Problem 6.}$ Let $P(z)$ be a polynomial of degree $n$ with complex co-efficients, $$P(0)=1, \quad \ \text{and} \ |P(z)| \leq M \ \ \text{for} \ |z| \leq 1.$$ Prove that every root of $P(z)$ in the closed unit disc has multiplicity at most $c\sqrt{n}$ where $c=c(M) > 0$ is a constant depending only on $M$.
$\textbf{Solution 1.}$ (As given in "Contests in Higher Mathematics" by : Gabor J.Szekely). It is sufficient to examine the multiplicity of number $1$. Infact, if we prove something for $1$ then we may appy the result to the polynomial $p(z)=P(\alpha z)$ with $|\alpha| \leq 1$, and in this way, we obtain the same estimate for all roots lying in the unit disc. The idea of the solution is the following. We consider the integral $$F(P) = \int\limits_{0}^{2\pi} \log{|P(e^{i\phi})|} \ d\phi$$ and show that it exists and is non-negative. Then we estimate it from above, once in the neighborhood of $1$ with the aid of multiplicity of $1$ and the degree of $P$, and once at other points using the condition $|P(z)| \leq M$.
It is sufficient to prove the existence of the integral for polynomials of the form $z-z_{0}$. If $$P(z)= c \cdot\prod\limits_{i=1}^{n}(z-z_{i})$$ then $$\log{|P(z)|} = \log{|c|} + \sum\limits_{i=1}^{n} \log{|z-z_{i}|}$$
The existence of $$\int\limits_{0}^{2\pi} \log{|e^{i\phi}-z_{0}|}\ d\phi$$ is evident if $|z_{0}| \neq 1$. Next let $|z_0|=1$. Without loss of generality, we may assume that $z_{0}=1$ (a substitution $\phi=n+\phi_{0}$ takes them into each other). Then $$|e^{i\phi}-1| = \left\lvert\:2\sin\frac{\phi}{2}\right\rvert$$ and $$\int\limits_{0}^{2\pi} \log\:\left\lvert\:2\sin\frac{\phi}{2}\right\rvert \ d\phi$$ really exists and is equal to $0$.
Next, compute the integral $$f(\alpha)= \int\limits_{0}^{2\pi} \log\biggl|1-\frac{e^{i\phi}}{\alpha}\biggr| \ d\phi \qquad\qquad (\alpha\neq 0)$$ for $\alpha \neq 1$. Obviously, its value depends on the absolute value of $\alpha$ only (again, a subsitution as above may be applied), so it is the same for the numbers $\alpha\epsilon_{1},\alpha\epsilon_{2},\cdots, \alpha\epsilon_{n}$ where the $\epsilon_{j}$ are the $n$-th unit roots. Therefore, $$n f(\alpha) = \sum\limits_{j=1}^{n} f(\alpha\epsilon_{j})=\int\limits_{0}^{2\pi} \log\:\Biggl| \prod\limits_{j=1}^{n} \biggl(1-\frac{e^{i\phi}}{\alpha\epsilon_{j}}\:\biggr)\Biggr| \ d\phi$$
Now if $|\alpha|>1$, then for $n\to\infty$ the integral on the right hand side tends to zero since $1-e^{in\phi}/\alpha^{n} \to 1$ uniformly; thus $f(\alpha)=0$. On, the other hand if $|\alpha| <1$ then $$n(f(\alpha)+2\pi\log{|\alpha|})=\int\limits_{0}^{2\pi} \log{|\alpha^{n}-e^{in\phi}|} \ d\phi \to 0$$ since $1-|\alpha|^{n} < |\alpha^{n}-e^{in\phi}|<1+|\alpha|^{n}$; so in this case $f(\alpha)=-2\pi \log{|\alpha|}$ is only possible. In each of the three case, we have $f(\alpha) \geq 0$. In our case, the relation $P(0)=1$ implies, $$P(z)=\prod\limits_{j=1}^{n} \biggl(1-\frac{z}{z_j}\biggr)$$ whence $$F(P)= \sum\limits_{j=1}^{n} f(z_j) \geq 0 \qquad (1)$$
Now let $P(z)=(z-1)^{k} Q(z)=a_{0}+a_{1}z+\cdots +a_{n}z^{n}$, where $Q(1)\neq 0$. We estimate $F(P)$ with the help of $k$, $n$, and $M$. Let $$F(P)=\int\limits_{0}^{2\pi} = \int\limits_{-\epsilon}^{\epsilon} + \int\limits_{\epsilon}^{2\pi-\epsilon} = F_{1}+F_{2}$$
Then $$F_{2} \leq \int\limits_{0}^{2\pi} \log{M} \ d\phi = 2 \cdot \pi \cdot \log{M} \qquad (2)$$
We split $F_{1}$ again into two parts: $$F_{1} = \int\limits_{-\epsilon}^{\epsilon} \log{|(z-1)^{k}|} \ d\phi + \int\limits_{-\epsilon}^{\epsilon} \log|Q(e^{i\phi})| \ d\phi=F_{3}+F_{4} \qquad (3)$$
Clearly
\begin{align*}
F_{3} &= k \int\limits_{-\epsilon}^{\epsilon} \log\Bigl(2\cdot\sin\frac{|\phi|}{2}\Bigr) < 2k\int\limits_{0}^{\epsilon} \log\phi \ d\phi \\ &=2k\epsilon \cdot (\log\epsilon -1) \qquad (4)
\end{align*}
For estimating $F_{4}$ we need an estimate of $Q$, which we obtain from the co-efficients of the expansion of $Q$ about $1$. Let $Q(1+z)= R(z)$. We calculate the co-efficients of $R$ from those of $P$ using the formula
$R(z)= \frac{P(z+1)}{z^{k}}$
\begin{align*}
P(z+1) &=\sum\limits_{j=0}^{n} a_{j}(z+1)^{j} =\sum\limits_{j=0}^{n}\sum\limits_{m=0}^{j}a_{j}\cdot {j \choose m} \cdot z^{m} \\ &=\sum\limits_{m=k}^{n} z^{m} \cdot \sum\limits_{j=m}^{n} a_{j} \cdot {j \choose m}
\end{align*}
since for $m < k$, the coefficient of $z^{m}$ is $0$ by our assumption. Thus $$R(z)=\sum\limits_{m=0}^{n-k}b_{m}z^{m}, \qquad b_{m}=\sum\limits_{j=m+k}a_{j} {j \choose m+k} \qquad (5)$$
Further, by the cauchy inequalities, $|a_{j}|=|P^{(j)}(0)|/j! \leq M$ Putting this into $(5)$, we find, $$|b_{m}| \leq M \sum\limits_{j=m+k}^{n} {j \choose m+k} = M {n+1 \choose m+k+1} \qquad (6)$$
If $|z|=\delta$ and $\delta(n-k)/(k+2) < 1$, then in view of $(6)$,
\begin{align*}
\frac{|R(z)|}{M} & \leq \sum\limits_{m=0}^{\infty} \delta^{m} {n+1 \choose m+k+1} \\ &={n+1 \choose k+1} \cdot \Bigl(1+\delta \frac{n-k}{k+2} + \delta^{2}\frac{n-k}{k+2}\cdot\frac{n-k-1}{k+3} + \cdots \Bigr) \\ &\leq {n+1 \choose k+1} \sum\limits_{j=0}^{\infty} \Bigl(\delta \frac{n-2}{k+2}\Bigr)^{j} = {n+1 \choose k+1} \frac{1}{1-\delta \frac{n-k}{k+2}}
\end{align*}
Since $|e^{i\phi}-1|=2|\sin\frac{\phi}{2}| \leq |\phi|$, therefore if $\epsilon < ((k+2)/2(n-k))$, then for $|\phi| \leq \epsilon$ we have $$|Q(e^{i\phi})|= |R(e^{i\phi}-1)| \leq M {n+1 \choose k+1}\frac{1}{1-\epsilon\frac{n-k}{k+2}} < 2M {n+1 \choose k+1}$$ If $k\geq 2$, $t! > t^{t}e^{-t}$ we obtain,
\begin{align*}
{n+1 \choose k+1} &= \frac{(n+1) \cdot n \cdot (n-1) \cdots (n-k+1)}{(k+1)!} \\ &= \frac{(n^{2}-1)\cdot n \cdot (n-2) \cdot (n-k+1)}{(k+1)!} < \frac{n^{k+1}}{(k+1)!} < \Bigl(\frac{en}{k+1}\Bigr)^{k+1}
\end{align*}
Thus $$\log{|Q(e^{i\phi})|} < \log{2M} + (k+1)\log\frac{en}{k+1}$$ and consequently $$F_{4} < 2\epsilon \biggl[ \log(2M) + (k+1)\log\frac{en}{k+1}\biggr]$$
Collecting everything, by $(1)$,$(2)$,$(3)$ and $(4)$ $$(\pi+\epsilon)\log{M} + \epsilon\log\frac{2en}{k+1} + \epsilon k \log\frac{en}{k+1} \geq 0 \qquad (7)$$
Now if $n=k^{2}/2c$, let $\epsilon = c/k$ (this fulfills the condition $\epsilon < ((k+2)/2(n-k))$). Then $(7)$ becomes, $$\Bigl(\pi+\frac{c}{k}\Bigr) \log{M} + \frac{c}{k} \log\frac{\epsilon k^{2}}{c(k+1)} + c \log\frac{k}{2(k+1)} \geq 0$$
We only make things worse if we also replace $k+1$ by $k$. Moreover, $c/k=k/2n \leq \frac{1}{2}$ gives $\pi+\frac{c}{k} < 4$, $c/k \log(ek/c) < (c/k) \cdot (ek/c)=e < 3$, So finally $$4\log{M} + 3 \geq c \log{2}, \qquad (8)$$ $$c < \log{M}+6$$ Since $k=\sqrt{2cn}$, relation $(8)$ means that we have proved the assertion of the problem with $$c(M)=\sqrt{16\log{M}+12}$$
