I have a function
$n \geq 2$
$h(z) = f(z)+g(z) = z^{n+1}-1+z-1$
$f(z)=z^{n+1}-1$
$g(z) = z-1$
is my proof is correct here By Using Rouché for $|z|\leq 1$
$|z^{n+1}-1| \leq |z|^{n+1}-1 \leq 0 \leq |z-1|$
meaning g(z) dominate f(z)
since g(z) has one root in $|z| \leq 1$ so does $f(z)$ is my proof correct thx
Proof of the fact that $f+g$ has exactly $n$ roots in $|z|>1$: Consider the disk of radius $1-\epsilon$ around then origin. Since $|z^{n+1}|=(1-\epsilon)^{n}< 1 <2-|z|\le |z-2|$ we see (by Rouchee's Theorem) that $f+g$ has no root in this disk. It follows that $f+g$ has no root in $|z|<1$. It remains to show that it has exctly $1$ root on $|z|=1$. If $|z|=1$ and $z^{n+1}+z-2=0$ we get $1=|z^{n+1}|=|2-z|\geq 2-|z|=1$ which forces $z$ to be $1$ (since the cirlces of radius $1$ around $0$ and $2$ intersect exactly at one point $z=1$).