Finding a disk containing all roots of a complex polynomial

I'm trying to list all roots of a polynomial so I found this paper, in Part 9 on page 29 it gives a simple recipe to find all the roots. But there is this remark:

We have assumed throughout the paper that our polynomial $p$ has all its roots in the unit disk. If instead the roots are in the disk $|z| < r$, you may simply scale all the starting points by a factor of $r$.

My problem is finding some $r$ which matches this. I can see that $r$ is basically any number larger than the maximum magnitude of all the roots. It dosen't matter if it's the smallest $r$, but the smaller the better.

I found a way that I have ben unable to find counter exambles to, albeit the $r$ will be much bigger in some cases, which is having $r$ be the product of the absolute value of all coefficients. (Not counting the polynomial $0$ since anything is a root). But I have no idea how I would go about proving that this $r$ is sufficient.

• Taking $r$ to be the product of the coefficients doesn't work; in fact, the product of the coefficients can be made arbitrarily small without changing the roots. For instance, $0.01 z^2 - 1$ has the same roots as $z^2 - 100$: $z=\pm 10$. – mjqxxxx May 15 '14 at 17:15
• See also this question. – Antonio Vargas May 15 '14 at 18:14
• See en.wikipedia.org/wiki/Properties_of_polynomial_roots for various root bounds. – LutzL May 19 '14 at 6:52

1 Answer

Eventually the leading-order term dominates. More precisely, if your polynomial is $P(z)=a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$, then for large enough $|z|$ the term $a_n z^n$ is greater in magnitude than the sum of the magnitudes of all the other terms, and there can be no more roots. We want $$|a_n z^n|>|a_{n-1}z^{n-1}|+\cdots + |a_1 z| + |a_0|;$$ dividing both sides by $|z|^n$ gives the equivalent $$|a_n|>|a_{n-1}/z|+\cdots + |a_1/z^{n-1}| + |a_0/z^n|,$$ which is clearly true as long as $|z|>1$ and $$|z|>\frac{|a_{n-1}|+\cdots+|a_1|+|a_0|}{|a_n|}.$$

• This is the Lagrange bound. Another is the Cauchy bound $r=1+\max\left\{\dfrac{|a_k|}{|a_n|}:k=0,...,n-1\right\}$. – LutzL May 19 '14 at 6:54