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?
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.