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.

  • $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 22:45

2 Answers 2


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

  • $\begingroup$ 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, 2017 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? $\endgroup$ Apr 22, 2017 at 1:06
  • $\begingroup$ I was wrong, I mixed up $h$ & $g$. $\endgroup$
    – copper.hat
    Apr 22, 2017 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, 2020 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 '$z^2 = 1$' $\endgroup$ Jan 19, 2020 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...


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