I understand that this means f has no roots and the polynomial is of degree zero. But why is that the case and how do we know that being greater than equal to 1 means f(x) has no roots?

also, are there any ways to analyse this apart from Liouville's Theorem (as I am not very acquainted with that)?

  • 2
    $\begingroup$ If $x$ is a root then $|f(x)|=0 <1$ $\endgroup$
    – Martin R
    Commented Apr 17, 2021 at 15:43
  • 2
    $\begingroup$ Since $1/f$ is a bounded entire function, it must be constant by Liouville's Theorem. $\endgroup$
    – S.C.B.
    Commented Apr 17, 2021 at 15:46
  • 1
    $\begingroup$ All polynomials of any degree greater than zero have at least one root in $\mathbb C$. This is the fundamental theorem of algebra. No roots means the degree must be less than 1.. $\endgroup$
    – David P
    Commented Apr 17, 2021 at 15:51
  • $\begingroup$ It looks like you are having a complex (analysis) day? :) $\endgroup$
    – rtybase
    Commented Apr 17, 2021 at 15:56
  • $\begingroup$ @rtybase yep! struggling with the concept :( $\endgroup$
    – user915841
    Commented Apr 17, 2021 at 15:59

1 Answer 1


As pointed out in the comments, what you need here is the fundamental theorem of algebra, which says that every nonconstant polynomial over $\mathbb{C}$ has a root in $\mathbb{C}$.

So, suppose that $f$ is a polynomial that satisfies $\lvert f(z) \rvert \geq 1$ for all $z \in \mathbb{C}$. If $f$ is nonconstant, then by the fundamental theorem of algebra there is a point $z_0$ such that $f(z_0) = 0$. But then $\lvert f(z_0) \rvert = 0 < 1$, which is a contradiction. Hence, any such $f$ must be a constant polynomial. (Indeed, if $a$ is any complex number such that $\lvert a \rvert \geq 1$, then the constant polynomial $f(z) = a$ satisfies the given condition.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .