5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $4$...) $\endgroup$ – Chappers Apr 21 '17 at 22:30
  • $\begingroup$ @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. $\endgroup$ – Mike Apr 21 '17 at 22:36
  • $\begingroup$ Simple example: if $|z|=1$, so $z=e^{i\theta}$, $|z+1| = \sqrt{(\cos{\theta}+1)^2+\sin^2{\theta}} = \sqrt{2+2\cos{\theta}} \neq |1+1|=2$. $\endgroup$ – Chappers Apr 21 '17 at 22:39
  • $\begingroup$ @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. $\endgroup$ – Mike Apr 21 '17 at 22:45
4
$\begingroup$

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 $(0, \pm\sqrt\frac{3}{5})$

$\endgroup$
10
  • $\begingroup$ Rouché's theorem doesn't apply if $f,g$ have a zero on the boundary of the region in question. $\endgroup$ – copper.hat Apr 22 '17 at 0:14
  • $\begingroup$ $f$ certainly can't (which is encoded in $f(z)| > |g(z)|$, but why can't $g$ have a zero on the boundary? $\endgroup$ – Badam Baplan Apr 22 '17 at 1:06
  • $\begingroup$ I was wrong, I mixed up $h$ & $g$. $\endgroup$ – copper.hat Apr 22 '17 at 5:09
  • $\begingroup$ $z^2$ is real also when $z=\pm i$ where $|f(z)|=8$ (which is not minimum of $f$), $f$'s minimum is at $\pm 1$ $\endgroup$ – Eran Jan 18 '20 at 19:58
  • $\begingroup$ @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 '$z^2 = 1$' $\endgroup$ – Badam Baplan Jan 19 '20 at 4:38
0
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.