Find the zeros of $h(z)=z^6-5z^4+3z^2-1$ within the unit disc - Verification Let $h(z)=z^6-5z^4+3z^2-1$.
Using Rouche's theorem, with $f(z)=-5z^4$ and $g(z)=z^6+3z^2-1$. 
On the unit disc $\lvert f(z) \rvert =5 > \lvert 1+3-1 \rvert=\lvert g(z)\rvert $
And the number of zeros is 4 for $f(z)$ in the unit disc, so it is also five for $h(z)=f(z)+g(z)$.
Is this correct? Thanks.
Found an old answer of mine, I chose $g(z)=z^6-1$ and $f(z)=-5z^4+3z^2$, and got the answer $4$ roots again.
 A: Your first approach with $f(z) = -5z^4$ and $g(z) = z^6 + 3z^2 - 1$ doesn't work, because $g(i) = -1 - 3 - 1 = - 5$, hence $|f(i)| = |g(i)|$
So you don't have a strict inequality on the boundary of the region.  True that $|f(z)| >= |g(z)|$, but that's not enough for Rouche's theorem.  
On the exam, your choice of functions does work for applying Rouche's theorem!  Taking $f(z) = -5z^4 + 3z^2$ and $g(z) = z^6 -1$, we now have that $|f(z)|$ takes its minimum of $2$ when and only when $z^2 = 1$, but at those points $|g(z)|$ is also at its minimum of $0$.  Meanwhile $|g(z)|$ takes its maximum at $2$, but this never happens when $z^2 = 1$.  We conclude that $|f(z)| > |g(z)|$ everywhere.
And $f(z)$ does have all four of its roots in the unit circle $(0, \pm\sqrt\frac{3}{5})$
A: This can be done by Rouche's theorem if you choose $g$ correctly. For future reference, there exists a stronger form of Rouche's theorem, which can be easier to apply. Loosely speaking, most books say this:


Rouche: If $|f-g|<|g|$ on the boundary then $f$ and $g$ have the same number of zeroes inside.


I don't know why they say that, because the following is no harder to prove:


New and Improved Rouche: If $|f-g|<|f|+|g|$ on the boundary then $f$ and $g$ have the same number of zeroes inside.


See for example Complex Made Simple...
