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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 Rouche's Theorem, where Rouche's theorem was used to show that the polynomial $p(z)=z^5+10z-3$ has $1$ zero in $\{z:\mathbb{C}:|z|<1\}$.

What was done:

Let $f(z)=10z-3$ and $g(z)=z^5+10z-3$. Let $|z|=1$. Then we get that:

$|f(z)-g(z)|\\=|-z^5|= |z^5|=1<z\le|10z-3|$

Hence $p(z)$ has 1 zero inside $\{z:\mathbb{C}:|z|<1\}$.

How did they get that $1<z$? What actually is $z$?

Secondly, why did they choose $f(z)=10z-3$? Can I choose $f(z)=10z$ and is what I did below correct?


Hence p(z) has 1 zero inside $\{z:\mathbb{C}:|z|<1\}$.

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In the first case, I think there's just a missing absolute value sign around the $z$. It should be $|z|$ instead. – Adelaide Dokras May 25 '12 at 20:45
Ahh okay, thanks for pointing it out to me! – Derrick May 25 '12 at 21:08
up vote 2 down vote accepted

So the only thing that concerns me about your formulation is that you assumed $|f(z)-g(z)|\leq |f(z)|$, rather than showing it. If we try to work backwards, starting with $4<10$ (the inequality needs to be strict), then we can recover $|-z^5| + |3| < 10|z|$ without a problem. Then we can use the triangle inequality because $|-z^5 + 3|\leq |-z^5|+|3|$, which recovers the step above. From there, we just fill in the missing terms to get the desired result $|f(z)-g(z)| < |f(z)|$.

Therefore, you can apply Rouché's Theorem. So you did it "right" but backwards. The reason they chose $10z-3$ instead is probably just so they didn't have to deal with the triangle inequality.

Edit: And yes, I agree with Adeal about $|z|$.

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Okay, thanks for your help and for pointing out my mistakes to me, most appreciated. – Derrick May 25 '12 at 21:07

Take $f(z)=10z-3,g(z)=z^5$, so on $|z|=1, |f(z)|\ge 7>|g(z)|=1$ so $f$ and $f+g=p$ has same number of zeros inside $|z|=1$

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I like this way. It is easier to remember. $|f(z)|>|g(z)|$ implies $f$ and $f+g$ (the bigger one i.e. $f$ appears twice! )have the same number of zeros. – TCL May 9 '13 at 14:28

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