Question : How many zeroes the polynomial $f(z)=z^4-5z+1$ are there in the disk $D(0; \frac{1}{4})$?
Rouché's theorem : Let $D \subset \mathbb{C}$ a domaine and $f,g: D \to \mathbb{C}$ two holomorphic functions in $D$. Let $C$ a closed path contained in the interior of $D$. If $|f(z)+g(z)| < |f(z)|+|g(z)|$, $\forall z \in \mathbb{C}$, then $f$ and $g$ have the same number of zeroes in the interior of $C$.
Let the function $g(z)=-z^{\alpha}$, with $\alpha \in \{0, \dots , 4\}$. I thought I could use this stategy to discover the good $\alpha$, but I blocked : $|f(z)+g(z)|= |z^4- z^{\alpha}+5(z)+1| \leq (\frac{1}{4})^4-(\frac{1}{4})^{\alpha}-5(\frac{1}{4})+1= \frac{64}{256}+(\frac{1}{4})^{\alpha}$
Are there possibilities to obtain something from this inequality? Otherwise, is anyone could give me a strategy to solve this problem? What is better to use here : Argument principal or Rouché's theorem?
Thanks
$$f(z) = 5z \implies M = |f(z)|_{\text{max}} = 5/4$$ $$g(z) = z^4+1\implies |g(z)|_{\text{max}} \le 1+\frac{1}{4^4} < M$$
Thus there is one solution, since $f$ having one root in $\{|z| \le 1/4 |z \in \mathbb C\}$ implies $f+g$ does as well. To see why this is, see the formulation of pg. 8: http://www.math.umn.edu/~bahra004/complex_prelim.pdf