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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 $|z|<1$.2011-12-23
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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.2011-12-23
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    Hey why don't you use the first two terms for |z|<2 ?!?2011-12-23
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    If you use the first two term, then you need to show that $|-4z+1|\leq |z^4+z^3|$ when $|z|=2$.2011-12-23

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