For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$

Solution 1:

Let me prove a slightly stronger property than the one you ask about since this can be done in one go:

Let $p(z) = a_{n} z^n + a_{n-1} z^{n-1} + \cdots + a_{0}$ be a complex polynomial with $a_{n} \neq 0$. There exists $R \gt 0$ such that $\frac{1}{2}|a_{n}|\,|z|^n \leq |p(z)| \leq \frac{3}{2} |a_{n}|\,|z|^n$ for all $z$ with $|z| \geq R$.

This is clearly true if $n = 0$, so let's assume $n \geq 1$. Put $q(z) = |a_{n-1}|\,|z|^{n-1} + \cdots + |a_0|$. By the triangle inequality we have the estimates $$|a_{n}|\,|z|^n - q(z) \leq |p(z)| \leq |a_{n}|\,|z|^n + q(z) \qquad \text{for all } z \in \mathbb{C}.$$ For $k \leq n-1$ we have $|z|^{k} \leq |z|^{n-1}$ if $|z| \geq 1$, so $$q(z) \leq (|a_{n-1}| + \cdots + |a_{0}|) |z|^{n-1} \qquad \text{for all } |z| \geq 1.$$ Putting $$R = \max{\left\{1, \frac{2}{|a_{n}|}(|a_{n-1}| + \cdots + |a_{0}|)\right\}}$$ we get the desired $R$ since for $|z| \geq R$ we then have $q(z) \leq q(z) \cdot \frac{|z|}{R} \leq \frac{|a_{n}|}{2} |z|^{n}$.

Solution 2:

Thanks for Qiaochu Yuan's hint. The solution:

Let $p(z) = q(z) + a_nz^n$.

Lemma: Let $c\neq 0$. $|cz^n| > |q(z)|$ for all $z\in \{z : |z|\geq r_c\}$, where $r_c$ is sufficiently large.

Proof: If $|z|=r$, then $|cz^n| = |c|r^n$. $$|q(z)| = |\sum_{i=0}^{n-1} a_iz^i| \leq \sum_{i=0}^{n-1}|a_iz^i| = \sum_{i=0}^{n-1}|a_i|r^i \leq \sum_{i=0}^{n-1} \max(|a_i|) r^i =\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i$$ when $r > 1$, $$\max_{i=0}^{n-1}(|a_i|) \sum_{i=0}^{n-1} r^i \leq \max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1}$$

If $r > \max(\frac{\max_{i=0}^{n-1}(|a_i|)(n-1)}{|c|},1)$ $$\max_{i=0}^{n-1}(|a_i|) (n-1) r^{n-1} \leq |c|r^n$$

$|p(z)| = |a_n z^n+q(z)|\geq |a_n z^n| - |q(z)| \geq |a_n z^n| - |c z^n| = (|a_n|-|c|)|z^n|$

Let $c = |\frac{a_n}{2}|$. For all $z$ such that $|z|=R>r_{a_n/2}$, this shows $|p(z)| \geq |a_n|R^n/2$.