When will maximum of the product=product of the maximum?
Clash Royale CLAN TAG#URR8PPP
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If $P(z)=prod_k=1^n(z-z_k),$ it is true in general that $max_ prod_k=1^n|z-z_k|neq prod_k=1^n(1+|z_k|).$ Simple example is $P(z)=(z-1)(z+1).$ Can we classify such polynomials satisfying $max_ prod_k=1^n|z-z_k|= prod_k=1^n(max_|z-z_k|)?$
complex-analysis
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If $P(z)=prod_k=1^n(z-z_k),$ it is true in general that $max_ prod_k=1^n|z-z_k|neq prod_k=1^n(1+|z_k|).$ Simple example is $P(z)=(z-1)(z+1).$ Can we classify such polynomials satisfying $max_ prod_k=1^n|z-z_k|= prod_k=1^n(max_|z-z_k|)?$
complex-analysis
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up vote
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down vote
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up vote
0
down vote
favorite
If $P(z)=prod_k=1^n(z-z_k),$ it is true in general that $max_ prod_k=1^n|z-z_k|neq prod_k=1^n(1+|z_k|).$ Simple example is $P(z)=(z-1)(z+1).$ Can we classify such polynomials satisfying $max_ prod_k=1^n|z-z_k|= prod_k=1^n(max_|z-z_k|)?$
complex-analysis
If $P(z)=prod_k=1^n(z-z_k),$ it is true in general that $max_ prod_k=1^n|z-z_k|neq prod_k=1^n(1+|z_k|).$ Simple example is $P(z)=(z-1)(z+1).$ Can we classify such polynomials satisfying $max_ prod_k=1^n|z-z_k|= prod_k=1^n(max_|z-z_k|)?$
complex-analysis
asked Aug 4 at 1:24
user159888
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