# Determine the number of zeros $\frac{1}{4} z ^{6} + z^{4}-z^{3}+1$ has inside the unit circle

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

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?

• 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. – Brian Moehring Apr 18 at 4:47
• 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. – copper.hat Apr 18 at 5:01
• Ohhh yes yes I thought the only use of this graph is to get the number of zero points of $p(z)$ – YiPing Apr 18 at 5:03
• Thanks you all guys! – YiPing Apr 18 at 5:03

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.