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This is a theorem that I have encountered before and proved using some homotopy-theoretic arguments, but I tried to prove this statement again without such tools and ended up failing. It seems to be Rouche's Theorem, but I can't figure out how to jimmy it just right.

The statement is very nice:

All roots of $p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0$ lie in the disc centered at zero with radius

$R = \sqrt{1 + |a_{n-1}|^2 + |a_{n-2}|^2 + \cdots + |a_0|^2}$

I keep thinking I have done it correctly, only to find an error. Does anyone have a slick proof of this that just uses complex-analytic techniques? I would really like to see just a sketch, or just the first step to start it off, but hey, I won't stop you from finishing it up haha

For those who would think it's useful to see the homotopy-theoretic version, check Munkres Topology, Ch. 9.56 and the first exercise, but it seems impossible to state this proof without using homotopy.

Thanks a lot!

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Let $|z|>1,$ and by Cauchy-Schwarz inequity \begin{align} |p(z)|&=|z|^n\left|1+\sum_{j=1}^{n}\frac{a_{n-j}}{z^j} \right|\geq |z|^n\left(1-\left|\sum_{j=1}^{n}\frac{a_{n-j}}{z^j}\right| \right)\\&\geq |z|^n\left[1-\left(\sum_{j=1}^{n}|a_{n-j}|^2\right)^{1/2}\left(\sum_{j=1}^{n}\frac{1}{|z|^{2j}} \right)^{1/2}\right]\\&>|z|^n\left[1-\left(\sum_{j=1}^{n}|a_{n-j}|^2\right)^{1/2}\left(\sum_{j=1}^{\infty}\frac{1}{|z|^{2j}} \right)^{1/2}\right]\\&=|z|^n\left[1-\left(\sum_{j=1}^{n}|a_{n-j}|^2\right)^{1/2}\left(\frac{1}{|z|^2-1} \right)^{1/2} \right] \end{align} Therefore, $|P(z)|>0$ if $1-\left(\sum_{j=1}^{n}|a_{n-j}|^2\right)^{1/2}\left(\frac{1}{|z|^2-1} \right)^{1/2}\geq 0$ i.e., if $|z|\geq \sqrt{1+|a_{n-1}|^2+|a_{n-2}|^2+\ldots+|a_0|}=R.$

Thus, all those zeros whose modulus is greater than 1 lies in $|z|\leq R,$ and those zeros whose modulus is less than or equal to 1 already lie in $|z|\leq R.$

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  • $\begingroup$ Thanks man, you're a hero! $\endgroup$ – John Samples Nov 24 '15 at 5:40

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