# Rouché's Theorem for $p(z)=z^7-5z^3+12$

Let $f$ and $g$ be differentiable on a domain $D$ and suppose that $\gamma$ is a simple closed contour whose inside is contained in D.

If $|f(z)-g(z)|<|f(z)|$ for all $z$ on $\gamma$, then $f$ and $g$ have the same number of zeros inside $\gamma$ (counted including their order).

I was reading an example of application of Rouché's Theorem, where Rouché's theorem was used to show that the polynomial $p(z)=z^7-5z^3+12$ has $0$ roots in $\{z:\mathbb{C}:|z|<1\}$.

What was done was:

Let $g(z)=z^7-5z^3+12$ and let $f(z)=12$. Then for $|z|=1$,

$|f(z)-g(z)|=|z^7-5z^3| \\ \le|z|^7+5|z| \\=1+5\\=6<12=|f(z)|$

Hence by Rouché's Theorem $p(z)=z^7-5z^3+12$ has $7$ roots in $\{z:\mathbb{C}:|z|<2\}$.

I was wondering, what is the purpose to do the step $\le|z|^7+5|z|$? Can't I just jump straight from $|f(z)-g(z)|=|z^7-5z^3| \\=|1-5|\\=4<12=|f(z)|?$

Secondly, $|f(z)-g(z)|=|-z^7+5z^3|$, is there a reason why they used $|f(z)-g(z)|=|z^7-5z^3|$?

Thirdly, what does it mean by "(counted including their order)"? (From the definition above)

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For the first question, it isn't true that $z^7$ or $z^3$ equal $1$ on the boundary of the unit disk (consider, say, $z=i.$) For the second question, they are equivalent - the second looks a bit neater. Order in this context means multiplicity. – user17794 May 25 '12 at 18:28
Thanks tim, appreciated. :) – Derrick May 25 '12 at 19:41
I may be wrong, but is your conclusion correct: $p(z)$ has $7$ roots in $D_2=\{z\in\mathbb{C}\mid |z|<2\}$ ? From your statement of Rouche's theorem you should conclude only that $p(z)$ has $0$ roots in $D_1$ since $f(z)=12$ has $0$ roots in $D_1$. – poirot May 26 '15 at 13:27

## 1 Answer

With $|z|=1$, $|f(z)-g(z)|=|z^7-5z^3| = |z^4-5|$, however the last quantity is not equal to $5-1$, as you can take any point with $|z|=1$ (eg, take $z=e^{i\frac{\pi}{4}}$, then the value is $6$). The best upper bound is the one given.

There is no difference between the $|-z^7+5z^3|$ and $|z^7-5z^3|$, since it appears under the $|.|$ (ie, |w| = |-w|).

Order means multiplicity. $z^2$ has a zero of multiplicity $2$ at $0$.

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Thanks @copper.hat I understand point 2 and 3 well. So I get it now we have to take any point with with $|z|=1$. But if I choose $z=i$ then $|z|=1$. I get that $|z^7-5z^3|=|i^7-5i^3|=|-i+5i|=|4i|$? – Derrick May 25 '12 at 19:16
Just to clarify, so when we are trying to apply rouche's theorem, we should try to find the (best) upper bound of $|f(z)-g(z)|=|z^7-5z^3|$ for |z|=1? – Derrick May 25 '12 at 19:18
You need to ensure that $|f(z)-g(z)| < |f(z)|$ for all $z$ such that $|z|=1$, not just specific $z$. By choice, you have $|f(z)| = 12$, so you need to make sure that $|f(z)-g(z)| < 12$ for any $|z|=1$. You don't need the best upper bound, any upper bound will do, as long as it is less than $12$. In the above case, the triangle inequality makes it easy to see that $6$ is an upper bound. – copper.hat May 25 '12 at 19:26
Okay thanks again copper.hat! I think I got it now. Most appreciated! – Derrick May 25 '12 at 19:41
@User69127: Not sure this helps, but if you compare $p(z) = z^2$ and $p_\epsilon(z) = z(z-\epsilon)$ (with $0<\epsilon<1$) on $|z|=1$, you get $|p(z)-p_\epsilon(z)| = \epsilon|z| < |p(z)|$, hence $p$ and $p_\epsilon$ have the same number of zeros (2) inside the circle. If $\epsilon=0$, the multiplicity of the zero of $p_\epsilon$ is 2, otherwise one. The point is that this application of Rouche's theorem is 'insensitive' to multiplicity as such. – copper.hat Apr 1 '14 at 21:53