I want to count number of zeros of $z^7-5z+1$ for $\{z | 1 < |z| < 3 \}$. Using Rouché's theorem I found that the function has 1 zero ($0$) of multiplicity 1 in $\{z | 1 < |z|\}$ and 1 zero ($0$, the same!) of multiplicity 7 in $\{z ||z| < 3 \}$. And how many zeros do we finally have in $\{z | 1 < |z| < 3 \}$?
The main problem is how one can count the difference? Why can we just writy that there are 6 zeros because $7-1=6$ and don't pay attention that $0$ is not in $\{z | 1 < |z| < 3 \}$?
We first consider the disc $|z|<1+\epsilon$ with $\epsilon$ small positive quantity (thanks for @saulplatz's comment for this correction), and the 2 functions $g=z^7$ and $f=-5z+1$ since $|g|< 1+\epsilon<5(1-\epsilon) -1)<|f| $ on $|z|=1+\epsilon$ we have by Rouche's Theorem that $|f|$ and $|f+g|$ have the same number of zeros on the disc $|z|=1+\epsilon$- that is 1. For $|z|<3$ we take $f=z^7$ and $g=-5z+1$. Obviously we have $|g|<|f|=3^7$ on $|z|=3$ so $f$ and $f+g$ have the same number of zeros inside $|z|<3$ which is 7. We thus conclude that in the annulus $1<|z|<3$ the function $z^7-5z+1$ has 6 zeros.