1
$\begingroup$

The diagram shows the path traced out by the polynomial $p(z) = z^{4}-z^{3}+1$ as $z$ goes around the unit circle. enter image description here

The problem asks to look at the picture to determine the number of zeros of $\frac{1}{4} z ^{6} +p(z)$ has inside the unit circle.

By argument principle we can see $p(z)$ has 2 zeros inside the unit circle. It seems we need to use Rouche's Theorem to get the number of zeros of $\frac{1}{4} z ^{6} +p(z)$. But I cannot see how to apply Rouche's Theorem here. Can someone help me?

$\endgroup$
4
  • 1
    $\begingroup$ Can you see that, if $z$ is on the unit circle, then $\left|\frac{1}{4}z^6\right| = \frac{1}{4}$? And all the values on the path in the diagram you've shown have magnitude strictly greater than $1/4$? That's exactly the setup for Rouche's theorem to show that $\frac{1}{4}z^6 + p(z)$ and $p(z)$ have the same number of zeros inside the unit circle. $\endgroup$ Apr 18, 2021 at 4:47
  • $\begingroup$ If you draw a circle of radius ${1 \over 4}$ around the curve you can see that it still encircles the origin twice. This is essentially the proof of Rouché. $\endgroup$
    – copper.hat
    Apr 18, 2021 at 5:01
  • $\begingroup$ Ohhh yes yes I thought the only use of this graph is to get the number of zero points of $p(z)$ $\endgroup$
    – YiPing
    Apr 18, 2021 at 5:03
  • $\begingroup$ Thanks you all guys! $\endgroup$
    – YiPing
    Apr 18, 2021 at 5:03

1 Answer 1

2
$\begingroup$

Well, all you need to do is $|p(z)|$ is greater than $\frac{1}{4}$ on the unit circle. This is equivalent to showing that $$3 - 2 \cos \theta + 2 \cos 4\theta -2 \cos 3 \theta > \frac14.$$ The right hand side of that is $2(1-\cos \theta) + 2(1-\cos 3 \theta) - 2(1-\cos 4\theta) + 1,$ so you may be able to finish the computation using your favorite double angle formula.

$\endgroup$

You must log in to answer this question.

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