Let $A, B \in M_n(\mathbb{C})$ such that $$|\det(A+zB)|\leq1$$ for any $z\in\mathbb{C}$ with $|z|=1$. Prove that $$|\det(A)|\le1$$
I assumed that $|\det A|\ge1$. Then there exists $A^{-1}$ and $|\det(A^{-1})|\le1$. I multiplied the relation and obtained $$|\det(I_n+zBA^{-1})|\le|\det(A^{-1})|\le1$$ so $|\det(I_n+zC)|\le1$ for any $z$ with $|z|=1$, where $C=BA^{-1}$. I don't think that this is true but I don't know how to prove it.
The $|\varphi|$ of holomorphic map $\varphi(z)=\det(A+zB)$ on the ball $\{|z|<1\}$ attains its maximum at $|z_{0}|=1$, so $|\det (A)|=|\varphi(0)|\leq|\varphi(z_{0})|\leq 1$.