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The task is to find number of $ {z^4} + {z^3} - 4z + 1 = 0$ in the area $1 < \left| z \right| < 2$. (this task is in the Rouché's theorem paragraph)

I used this theorem many times, but I don't know to solve this task. This is simple to find number of roots in the area $0 < \left| z \right| < 1$, but I don't know how to do the same in another area: $0 < \left| z \right| < 2$.

Of course, I now number of roots with the help of Wolfram Alpha, for example.

Could you help, please, how to solve this task with the Rouché's theorem?

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Hint: The number of roots of $p(z)$ with $|z|<2$ are the number of roots of $p(2z)$ with $|z|<1$. –  Alexander Thumm Dec 23 '11 at 8:46
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@Alexander: Yes, but that doesn't get us all that far. The problem is that whereas for $|z|\lt1$ three of the coefficients are less than the fourth and can therefore be dropped, for $|z|\lt2$ we have $p(2z)=16z^4+8z^3-8z+1$, and now $8+8+1\gt16$, so it takes a bit more work to bound the absolute value to allow us to reduce this to $16z^4$ by Rouché's theorem. –  joriki Dec 23 '11 at 11:27
    
Hey why don't you use the first two terms for |z|<2 ?!? –  uncookedfalcon Dec 23 '11 at 12:49
    
If you use the first two term, then you need to show that $|-4z+1|\leq |z^4+z^3|$ when $|z|=2$. –  Paul Dec 23 '11 at 12:58
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1 Answer

Let's Recall the statement of the theorem:

Theorem: (Rouché) Let $A$ be some region enclosed by a simple loop $\gamma$. If $f(z)$ and $g(z)$ are analytic on an open neighborhood of $A$, and $|f(z)|>|g(z)|$ on $\gamma$, then $f(z)$ and $f(z)+g(z)$ have the same number of zeros inside $A$.

Circle |z|=1: Let $p(z)=z^4+z^3-4z+1$ be your polynomial. First lets count the number of zeros inside the disk $|z|=1$. Since $$|-4z|\geq 3\geq |z^4+z^3+1|,$$ we know that there will be one zero in this region by applying Rouché's Theorem with $f=-4z$.

Circle |z|=2: To find the number of zeros inside the circle $|z|=2$, we have to be slightly trickier. Notice that we have the miraculous identity

$$p(z)=\left(z^4+z^3-z^2-z\right)+\left(z^2-3z+1\right)=z(z-1)(z+1)^2+(z+1)^2-z.$$

With some manipulation of inequalities we also have the following lemma

Lemma: The inequality $$2|z-1||z+1|^2> |z+1|^2+2$$ holds for all $|z|=2$.

Consider $f(z)=z(z-1)(z+1)^2$ and $g(z)=(z+1)^2-z$ in Rouché's Theorem. When $|z|=2$, we have both $$|(z+1)^2|+2\geq |(z+1)^2-z|,$$ and $$|z(z-1)(z+1)^2|\geq 2|(z-1)(z+1)^2|.$$ Hence by the lemma $|f(z)|>|g(z)|$ for $|z|=2$ so that $p(z)$ has four roots inside the disk $|z|=2$.

Conclusion: As there are no zeros with $|z|=2$ or $|z|=1$, we conclude there are $3$ zeros in the region $1<|z|<2$.

Hope that helps,

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