# if $f(x)$ is a polynomial with complex coefficients, what does $|f(x)|\geq 1$ for all $x\in \mathbb{C}$ imply?

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)?

• If $x$ is a root then $|f(x)|=0 <1$ Commented Apr 17, 2021 at 15:43
• Since $1/f$ is a bounded entire function, it must be constant by Liouville's Theorem. Commented Apr 17, 2021 at 15:46
• 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.. Commented Apr 17, 2021 at 15:51
• It looks like you are having a complex (analysis) day? :) Commented Apr 17, 2021 at 15:56
• @rtybase yep! struggling with the concept :( Commented Apr 17, 2021 at 15:59

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.)