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Consider $z^6 - 6z^2 + 10z + 2$ on the annulus $1<|z|<2$.

By Rouche's Theorem $|f(z) + g(z)| < |f(z)|$ implies that both sides of the inequality have the same number of zeros. I understand that when asked to find that there is, say, $1$ zero the idea is to choose the $10z$ term as $f(z)$ so that we can form the proper relationship and conclude that since $10z$ has one zero in the region the function has one zero in the region. How can it be set up to find a general number of zeros? What is the trick for picking $f(z)$?

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Problems of this type which come from books are often vulnerable to some meta-mathematical arguments. One basic principle is that, in order to apply Rouché's theorem, you are never expected to count the number of zeros of any polynomial which you cannot solve explicitly. Also, in the actual application of Rouché's theorem, you'll rarely need to use anything more complicated than the triangle inequality to establish the required inequality. In practice this means that you're hoping to show something like

$$ |\text{complicated}| < |\text{simple}|. $$

Another guiding rule is that larger powers of $z$ are smaller in the unit circle and smaller powers of $z$ are smaller outside of the unit circle when compared to the rest of the polynomial.

For this particular problem, if I wanted to know how many zeros are inside the unit circle, I would guess that the dominating polynomial $q(z)$ in the inequality $|p(z)| < |q(z)|$ is either $2$ or $10z+2$ or $10z$, since smaller powers are more important in the unit circle. For convenience we'll try $10z$ first.

The polynomial $10z$ has one zero at $z=0$ which happens to lie inside the unit circle. Now on $|z| = 1$ we have

$$ \begin{align} \left|z^6-6z^2+2\right| &\leq 1+6+2 \\ &= 9 \\ &< 10 \\ &= 10 |z| \\ &= |10z|, \end{align} $$

and we can conclude from Rouché's theorem that $z^6 - 6z^2 + 10z + 2$ has exactly one zero in the disk $|z| \leq 1$. If you'd like you can check that it's also true that

$$ \left|z^6-6z^2\right| < |10z+2| $$

on the unit circle, so we could have gone that way instead.

To find the number of zeros in $|z| \leq 2$ we'll start from the other end and hope that $z^6$ dominates on the circle $|z| = 2$. Note that $z^6$ has a zero of multiplicity $6$ at $z=0$ (so, effectively, it has $6$ zeros), which happens to lie inside the circle $|z| = 2$. On $|z| = 2$ we have

$$ \begin{align} \left|-6z^2+10z+2\right| &\leq 6 \cdot 2^2 + 10 \cdot 2 + 2 \\ &= 46 \\ &< 64 \\ &= 2^6 \\ &= \left|z^6\right|, \end{align} $$

so from Rouché's theorem we know that $z^6 - 6z^2 + 10z + 2$ has all $6$ zeros in the disk $|z| < 2$.

Hence the number of zeros of the polynomial in the annulus $1 < |z| < 2$ is $6-1 = 5$.

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This is an excellent response, thank you. Was it simply guess work choosing $z^6$ and $z$ though for each circle? What if we had guessed there might be two roots in the circle of radius 1 with the rest in the circle of radius 2, or that in fact that in an unlikely case 5 of them lie within 2 and the other is somewhere far outside. – user73041 Apr 29 '13 at 18:51
@user73041, I wouldn't really call it guess work. Like I said in the beginning of the answer the larger powers of $z$ are more important outside the unit circle, so I expected $z^6$ to dominate on $|z| = 2$. The smaller powers of $z$ are more important inside the unit circle, so I expected $10z$, $10z+2$, or $2$ to dominate on $|z| = 1$. I'm not thinking about the possible locations of the zeros at all---I'm thinking about the relative sizes of the terms of the polynomial. (By "dominate" I mean that $b$ dominates $a$ if $|a| < |b|$.) – Antonio Vargas Apr 29 '13 at 18:55
Would it be possible though that, say in a case where you have $z^6$ and $z^7$ choose the larger and turn out to be wrong? That is, in fact 6 are in the circle of, say, radius 2 and the rest are somewhere outside (from 2 to infinity). – user73041 Apr 29 '13 at 19:02
@user73041, sure, the principle is only a rule of thumb. Picking the largest or the smallest won't always work but it's a good place to start. I had a good feeling about $z^6$ in particular because it's so much larger than the next term, $-6z^2$. – Antonio Vargas Apr 29 '13 at 19:05
That makes sense, thank you again. – user73041 Apr 29 '13 at 19:12

You can just split up the problem in two -- $|z|<1$ and $|z|<2$. Then you just subtract the number of zeros in the smaller regions from the zeroes of the bigger region, and you have your answer.

Both times you just apply Rouche's Theorem.

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Right, but my main concern is: for both such cases how does one determine f(z)? We have no knowledge of how many zeros might be in each region so far as I can tell. If the question had told us that we were looking for 1 zero in the region |z| <1 then I would expect to use 10z but without that information I am not sure how exactly one determines the 'trick' that allows us to invoke Rouche. I worry that even if I am able to come up with a trick here I could easily have no idea what the trick is for another problem. – user73041 Apr 28 '13 at 14:46

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