bounding roots of a polynomial with Rouche's Theorem Suppose f(z) = z^n - k [ z^(n-1) + ... + z + 1 ] where n is a positive integer and k is a real constant such that nk<1.
I have shown that a root of this polynomial must satisy |z|<1, but I want a slightly better bound such as 1-k. This seems plausible from computational results but is difficult to prove. I am trying to use Rouche's theorem to do this but finding an appropriate bounding function is difficult. Is there any other result about holomorphic functions that may help?
 A: Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|\ge f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. 
Differentiating $f(z)(z-1)$, one gets a bound:
$$z_k < \frac{1 + k}{1 + n^{-1}}.$$
When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.
The OP complains that he wants something better. It is pointed out that
 as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to 
$$\frac{2(1 - kn)}{(1 + n)}.$$
The OP then complains that he wants a bound in $n$ and $k$ (which was already given).
The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?"
No --- mathematics was used at this point.
The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.
It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time.
To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".
A: I don't know about any further boundings, but n = 3 and k = 1/4, or polynomial $4z^3 - z^2-z-1 = 0$ has a solution (1/12 + 1/12 (235 - 6 Sqrt[1473])^(1/3) + 
 1/12 (235 + 6 Sqrt[1473])^(1/3)), whose absolute value is ~ 0.868877, which is greater than 1-k.  Other {n,k} pairs are {2,3}, {4,6}, and {5,6}.
EDIT I noticed that if the roots are multiplied by nk, then as k goes from 0 to 1/n, the largest root in absolute value (which happens to be the largest root) goes from 0 to about 1.  So I suppose that the roots are bound in the range (0, 1/n).
