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Oscar Lanzi
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Answer became outdated when the question was edited. Originally $2M$ was stated as an upper not lower bound, due to a typo. Watch your signs!

Yup, as suggested by Fedor in the comments the premise is false. For example, $P(z)=z^4+z^3+z^2+z+1$ has all zeroes on the boundary of the unit disk and all coefficients unity, but $|P(1)|>2$.

If $z$ is inside or on the unti disk, then all polynomial functions (regardless of their zeroes) satisfy $|P(z)|\le\sum|a_n|$ from the Triangle Inequality.

Yup, as suggested by Fedor in the comments the premise is false. For example, $P(z)=z^4+z^3+z^2+z+1$ has all zeroes on the boundary of the unit disk and all coefficients unity, but $|P(1)|>2$.

If $z$ is inside or on the unti disk, then all polynomial functions (regardless of their zeroes) satisfy $|P(z)|\le\sum|a_n|$ from the Triangle Inequality.

Answer became outdated when the question was edited. Originally $2M$ was stated as an upper not lower bound, due to a typo. Watch your signs!

Yup, as suggested by Fedor in the comments the premise is false. For example, $P(z)=z^4+z^3+z^2+z+1$ has all zeroes on the boundary of the unit disk and all coefficients unity, but $|P(1)|>2$.

If $z$ is inside or on the unti disk, then all polynomial functions (regardless of their zeroes) satisfy $|P(z)|\le\sum|a_n|$ from the Triangle Inequality.

Source Link
Oscar Lanzi
  • 2.4k
  • 21
  • 20

Yup, as suggested by Fedor in the comments the premise is false. For example, $P(z)=z^4+z^3+z^2+z+1$ has all zeroes on the boundary of the unit disk and all coefficients unity, but $|P(1)|>2$.

If $z$ is inside or on the unti disk, then all polynomial functions (regardless of their zeroes) satisfy $|P(z)|\le\sum|a_n|$ from the Triangle Inequality.