# Find the zeros of $h(z)=z^6-5z^4+3z^2-1$ within the unit disc - Verification

Let $h(z)=z^6-5z^4+3z^2-1$.

Using Rouche's theorem, with $f(z)=-5z^4$ and $g(z)=z^6+3z^2-1$.

On the unit disc $\lvert f(z) \rvert =5 > \lvert 1+3-1 \rvert=\lvert g(z)\rvert$

And the number of zeros is 4 for $f(z)$ in the unit disc, so it is also five for $h(z)=f(z)+g(z)$.

Is this correct? Thanks.

Found an old answer of mine, I chose $g(z)=z^6-1$ and $f(z)=-5z^4+3z^2$, and got the answer $4$ roots again.

• You need to be more careful with $g$: its modulus is not constant, since it contains different powers. Moreover, $f$ has degree $4$, so it can't have $5$ zeros! (I think you mean $4$...) – Chappers Apr 21 '17 at 22:30
• @Chappers Thanks, corrected the 5 to a 4. I thought Rouche's theorem needs the modulus of $f$ to be greater on the boundary of the region, and there the powers are all powers of 1. – Mike Apr 21 '17 at 22:36
• Simple example: if $|z|=1$, so $z=e^{i\theta}$, $|z+1| = \sqrt{(\cos{\theta}+1)^2+\sin^2{\theta}} = \sqrt{2+2\cos{\theta}} \neq |1+1|=2$. – Chappers Apr 21 '17 at 22:39
• @Chappers That's interesting, obviously I can't just plug in $z=1$ since there are all the $e^{i \theta}$ choices. But I have no idea how to treat the modulus otherwise. I found my old exam with that answer on it, I will post it as a note, but there I did the same process with different choices of functions. The professor accepted it, I haven't seen any Rouche's Theorem applications do anything more elaborate with the functions on the boundary than that. – Mike Apr 21 '17 at 22:45

Your first approach with $$f(z) = -5z^4$$ and $$g(z) = z^6 + 3z^2 - 1$$ doesn't work, because $$g(i) = -1 - 3 - 1 = - 5$$, hence $$|f(i)| = |g(i)|$$

So you don't have a strict inequality on the boundary of the region. True that $$|f(z)| >= |g(z)|$$, but that's not enough for Rouche's theorem.

On the exam, your choice of functions does work for applying Rouche's theorem! Taking $$f(z) = -5z^4 + 3z^2$$ and $$g(z) = z^6 -1$$, we now have that $$|f(z)|$$ takes its minimum of $$2$$ when and only when $$z^2 = 1$$, but at those points $$|g(z)|$$ is also at its minimum of $$0$$. Meanwhile $$|g(z)|$$ takes its maximum at $$2$$, but this never happens when $$z^2 = 1$$. We conclude that $$|f(z)| > |g(z)|$$ everywhere.

And $$f(z)$$ does have all four of its roots in the unit circle $$(0, \pm\sqrt\frac{3}{5})$$

• Rouché's theorem doesn't apply if $f,g$ have a zero on the boundary of the region in question. – copper.hat Apr 22 '17 at 0:14
• $f$ certainly can't (which is encoded in $f(z)| > |g(z)|$, but why can't $g$ have a zero on the boundary? – Badam Baplan Apr 22 '17 at 1:06
• I was wrong, I mixed up $h$ & $g$. – copper.hat Apr 22 '17 at 5:09
• $z^2$ is real also when $z=\pm i$ where $|f(z)|=8$ (which is not minimum of $f$), $f$'s minimum is at $\pm 1$ – Eran Jan 18 '20 at 19:58
• @Yea you're right, I don't know why I wrote that, it's wrong. The reasoning still holds with '$z^2$ is real' replaced by '$z^2 = 1$' – Badam Baplan Jan 19 '20 at 4:38

This can be done by Rouche's theorem if you choose $g$ correctly. For future reference, there exists a stronger form of Rouche's theorem, which can be easier to apply. Loosely speaking, most books say this:

Rouche: If $|f-g|<|g|$ on the boundary then $f$ and $g$ have the same number of zeroes inside.

I don't know why they say that, because the following is no harder to prove:

New and Improved Rouche: If $|f-g|<|f|+|g|$ on the boundary then $f$ and $g$ have the same number of zeroes inside.

See for example Complex Made Simple...