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This is an example from a book.

Let $f(z) = z^5 + z^4 + 6z +1.$ Then since for $|z| = 2,$ $$|z^4 + 6Z + 1| \leq 29 < 32 = |z|^5$$ and for $|z| = 1,$ $$|z^5 + z^4 + 1| \leq 3 < 6 = |6z|.$$ By Rouche Theorem, $f$ has $5$ zeros in the disc $|z| < 2$ and $1$ zero in the disc $|z| < 1.$

Then it concludes that $f(z)$ $\textbf{has 5 - 1 = 4}$ aeros in the annulus $1 < |z| < 2.$

I think that from Rouche Theorem, $f$ should has 4 zeros in $1 \leq |z| < 2.$ For some obvious reasons (which is not stated), there is no zeros on $|z| = 1.$

Is there any clear ways to see why there are no zeros on $|z| =1 ?$

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    $\begingroup$ Your second displayed formula demonstrates the lack of zeroes on the unit circle quite clearly. $\endgroup$ – Lord Shark the Unknown May 17 '18 at 19:27
  • $\begingroup$ Ok, I guess by contradiction and the second inequality proves it. Is it always like this ? I mean apply Rouche to two circles with two radiuses, the difference of zeros is the number of zeros in the annulus (no need to worry about zeros on the circles). $\endgroup$ – Both Htob May 17 '18 at 19:42

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