1
$\begingroup$

I need to find the zeros of $z^6-4z^5+z^2-1$ in the unit disk using Rouche's theorem.

The hint given is to consider the functions $g(z)=z^6-4z^5+z^2-1$ and $f(z)=-4z^5$. The issue is that the inequality $|f(z)-g(z)|<|f(z)$, that is, $|z^6+z^2-1|<4|z^5|$ does not seem to hold on the unit disk. What am I doing wrong?

Also, in general, how do you find the function $g$ required to apply Rouche's theorem?

$\endgroup$
1
  • $\begingroup$ Yes. Perhaps I'm missing something obvious, but I still can't get the inequality. $\endgroup$ May 5, 2015 at 0:17

1 Answer 1

2
$\begingroup$

You have to check the inequality holds on the boundary. This is easy for a disk: $\lvert z \rvert =1$, so you have $$ \lvert z^6+z^2-1 \rvert < \lvert z^6 \rvert + \lvert z^2 \rvert + 1 = 3, $$ and $$ \lvert 4z^5 \rvert = 4>3. $$ Hence $ -4z^5 $ is larger in modulus than $z^6+z^2-1$ on the boundary of the unit disc, so $-4z^5$ and $(z^6+z^2-1)+(-4z^5)$ have the same number of roots inside.

In general, you can't tell. Here you look at the term with the largest coefficient, but there's no general way to tell; worse, it may not be possible to find such a splitting.

$\endgroup$
1
  • $\begingroup$ "On the boundary". Thanks a lot. $\endgroup$ May 5, 2015 at 0:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .