Use of Rouche's Theorem

Question

Good day, I am working on the question where $g(z) = z^4 - 2z - 2$ in the domain of $\frac{1}{2} < |z| < \frac{3}{2}$ I am trying to find the number of zeros within the annulus.

So by Rouche's Theorem, I have two ways to do this: either pick an $f$ to show that $|f-g| < |f|$ or Let $g(z) = s(z) + f(z)$ and for $|s|<|f|$, $f$ and $f+s$ have the same number of zeros.

So I tried to find the number of zeros for $|z| = \frac{1}2$ using both methods, and I get the same two answers with both methods. I can pick one function and I will see that there is one zero. But I can pick another function and I will get there is no zero.

For example, take $f = -2$ => there is no zero in $f\ |z|= \frac{1}2$ or take $f = -2z-2$ => there is one zero in $f$ within $|z|= \frac{1}2$

Thank you for your help.

Answer 1

By the fundamental theorem of Algebra, $g$ has four zeros in $\Bbb C$. If $|z| = 3/2$, we have: $$|2z+2| \leq 2|z|+2 = 5 = \frac{80}{16} < \frac{81}{16} = |z|^4,$$so all zeros are inside the region $\{ |z| < 3/2\}$. On the other hand, if $|z| = 1/2$, we have: $$|z^4-2z| \leq |z|^4 + 2|z| = \frac{1}{16}+1 < 2,$$so $g$ has no zeros in $\{|z| < 1/2\}$. We only have to check in the circle $|z| = 1/2$ now. Such $z$ would satisfy $z^4 - 2z-2 = 0$ and so: $$z^4 = 2z+2 \stackrel{|\cdot|}{\implies} \frac{1}{16} = |2z+2| \geq 2 - 2|z| = 1,$$contradiction. So all four zeros are in the annulus $\{1/2 < |z| < 3/2\}$.

Answer 2

So the question is why $$ |s(z)|=|z|^4=\frac1{16}<1=2-2|z|\le |2+2z|=|f(z)| $$ is in conflict with the claim that $f(z)=-2z-2$ and $g(z)=s(z)+f(z)=z^4-2z-2$ have the same number of roots, which is zero, in $D(0,\frac12)$.

But $f(z)$ has only one root $z=-1$ which is outside the disk, so also this line of inquiry shows that $g(z)$ has no root inside the disk.