$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $\mid z\mid =1$, we have
$\mid f(z)-g(z) \mid=1$ and $\mid f(z) \mid=4$
so $$\mid f(z)-g(z)\mid \leq \mid f(z)\mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?
Best Answer
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $\lvert f(z)-g(z)\rvert\ne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then $$ \lvert h(z)\rvert\le \lvert z\rvert^7+2\lvert z\rvert^5+\lvert z\rvert +1\le 5,$$ $$\lvert f(z)\rvert= \lvert 6z^3\rvert=6.$$ So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.