A little detail in a question about finding all zeros of $2z^5 +8z -1$ in the ring $\{ 1 < |z| < 2 \}$

Question

The question is to find the number of zeros of the function $f(z) = 2z^5 +8z -1$ in the ring $\{ 1 < |z| < 2 \}$. The answer to this question says that there are $4$.

So I used Roche's theorem in a standard way, and got to those results:

In $\{ |z|<1 \}$ there is one zero. In $\{ |z|<2 \}$ there are $5$ zeros.

So by subtracting the domains I get that there are $4$ zeros in $\{ 1 \le |z| < 2 \}$. What I don't understand is: why on the domain $\{ |z|=1 \}$ always $f(z) \neq 0 $.

Help would be appreciated.

Answer 1

Note that when $|z|=1$, we have $|f(z)| = |2z^5+8z-1| = |8z - (1-2z^5)| \ge |8z| - |1-2z^5| \ge 8 - 3 = 5$.

Answer 2

assume there is a solution $z_0$ on $|z|=1$ s.t f($z_0$)=0 than $2z_0^5 +8z_0-1=0$ therefore $2z_0^5=-8z_0+1$ so $|2z_0^5|=|8z_0-1|$ in contradiction for what you got by using the Rouche's theorem.