I have a polynomial $p$ of degree $n$ satisfying $\lvert p(z) \lvert \leq c\ \ \forall z\in\partial B_1(0)$. (Isn't this true for any polynomial?) Show $\lvert p(z)\lvert \leq c \lvert z\lvert^n \ \ \forall z\in \mathbb{C}\backslash B_1(0)$.

The obvious attempt would be $|p(z)|=|p(\lvert z\lvert\frac{z}{\lvert z\lvert})|=|\sum_{i=0}^n a_i |z|^i (\frac{z}{\lvert z\lvert})^i|\leq |z|^n \sum_{i=0}^n |a_i (\frac{z}{\lvert z\lvert})^i|$ which doesn't lead anywhere.

I guess I have to apply some maximum principle but don't know how.

  • $\begingroup$ Yes, use the maximum modulus principle. Note that $|p(z)|$ equals $|p(z)/z^n|$ on the boundary of the unit disk. $\endgroup$
    – hardmath
    Commented Jun 5, 2012 at 15:36

1 Answer 1


Let $\Omega:=\{z\in\Bbb C, |z|<1\}$. It's a domain. Define $g(z):=z^np\left(\frac 1z\right)$: it's a holomorphic function, and by maximum modulus principle the modulus is reached at the boundary, and is $\leq c$. Now, for $|z|>1$, work with $\frac 1z$ in the previous argument.

  • 1
    $\begingroup$ $\Omega$ is unbounded and the maximum modulus principle does not apply. Otherwise the argument in your first paragraph would apply to a function like $e^z$. $\endgroup$ Commented Jun 5, 2012 at 15:45
  • $\begingroup$ Indeed, it's a bit more complicated than that. I guess we can work with open $\Omega_R$ where $\Omega_R:=\{z,1<|z|<R\}$. I will try whether I can salvage this attempt. Thanks anyway! $\endgroup$ Commented Jun 5, 2012 at 15:49
  • $\begingroup$ Although $\Omega$ is unbounded, it (plus the point at infinity) is the image of $B_1$ under the Moebius mapping $z \rightarrow 1/z$. So apply the maximum modulus principle to the right polynomial on $B_1$ in $y = 1/z$. $\endgroup$
    – hardmath
    Commented Jun 5, 2012 at 19:35

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