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It has to be a non-constant polynomial by default and hence explicitly expressed as $n\geq 1.$
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Let $P(z)=\sum_{m=0}^na_mz^m$ be a polynomial of degree $n$$n\geq 1$ having no zeros in $|z|<1,$ then for any complex number $\alpha$ with $|\alpha|=1,$ is it true on $|z|=1$ that $$\left|\alpha zP'(z)+zP'(z)-\left(n-\frac{|a_0|-1}{|a_0|+1}\right)P(z)\right|\leq \left(n-\frac{|a_0|-1}{|a_0|+1}\right)\max_{|z|=1}|P(z)|?$$ The inequality arises from the case of $\alpha=-1.$

Let $P(z)=\sum_{m=0}^na_mz^m$ be a polynomial of degree $n$ having no zeros in $|z|<1,$ then for any complex number $\alpha$ with $|\alpha|=1,$ is it true on $|z|=1$ that $$\left|\alpha zP'(z)+zP'(z)-\left(n-\frac{|a_0|-1}{|a_0|+1}\right)P(z)\right|\leq \left(n-\frac{|a_0|-1}{|a_0|+1}\right)\max_{|z|=1}|P(z)|?$$ The inequality arises from the case of $\alpha=-1.$

Let $P(z)=\sum_{m=0}^na_mz^m$ be a polynomial of degree $n\geq 1$ having no zeros in $|z|<1,$ then for any complex number $\alpha$ with $|\alpha|=1,$ is it true on $|z|=1$ that $$\left|\alpha zP'(z)+zP'(z)-\left(n-\frac{|a_0|-1}{|a_0|+1}\right)P(z)\right|\leq \left(n-\frac{|a_0|-1}{|a_0|+1}\right)\max_{|z|=1}|P(z)|?$$ The inequality arises from the case of $\alpha=-1.$

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An inequality with rotation

Let $P(z)=\sum_{m=0}^na_mz^m$ be a polynomial of degree $n$ having no zeros in $|z|<1,$ then for any complex number $\alpha$ with $|\alpha|=1,$ is it true on $|z|=1$ that $$\left|\alpha zP'(z)+zP'(z)-\left(n-\frac{|a_0|-1}{|a_0|+1}\right)P(z)\right|\leq \left(n-\frac{|a_0|-1}{|a_0|+1}\right)\max_{|z|=1}|P(z)|?$$ The inequality arises from the case of $\alpha=-1.$