Find number of roots in some area (Rouché's theorem) 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?
 A: 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,
