Lower bound for complex polynomial beyond circle or radius R

Question

If we have a polynomial with $c_i$ a complex number $$c_nz^n + c_{n-1}z^{n-1} + \cdots + c_1 z + c_0$$ then $$|P(z)| > \frac{|c_n|R^n}{2}$$ When |z| > R for some R

I have tried using the triangle inequality where I obtain, $|P(Z)| \leq |c_n||z|^n + \cdots + |c_0|$. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!

Answer 1

You have $$\frac{|P(z)|}{|z|^n} =\left|c_n+\frac{c_{n-1}}z+\frac{c_{n-2}}{z^2}+\cdots+\frac{c_0}{z^n}\right|.$$ Show that if $|z|$ is large enough, then $$\left|\frac{c_{n-1}}z+\frac{c_{n-2}}{z^2}+\cdots+\frac{c_0}{z^n}\right|<\frac{|c_n|}2$$ etc.

Answer 2

This is part of Growth lemma, we have $$|p(z)|=|z^n(c_n +\dfrac{ c_{n-1}}{z} + \cdots + \dfrac{c_1}{z^{n-1}} + \dfrac{c_0}{z^n})|$$ for $|z|=r$ $$|p(z)|\geq r^n\left(|c_n|-|\dfrac{ c_{n-1}}{z} + \cdots + \dfrac{c_1}{z^{n-1}} + \dfrac{c_0}{z^n})|\right)$$ but $$|\dfrac{ c_{n-1}}{z} + \cdots + \dfrac{c_1}{z^{n-1}} + \dfrac{c_0}{z^n})|\leq\dfrac{1}{r}\left(|c_{n-1}+\cdots+|c_0|\right)=\dfrac{\alpha}{r}$$ therefore $$|p(z)|\geq r^n\left(|c_n|-\dfrac{\alpha}{r}\right)$$ now let $r>\dfrac{2}{|c_n|}\alpha$.