MHB Number of Zeros Inside Unit Circle for Polynomial P(z)

[Dustinsfl]

Let $P(z) = z^8 - 5z^3 + z - 2$. We want to find the number of roots of this polynomial inside the unit circle.
Let $f(z) = -5z^3$ (Why is this being chosen?)

Then $|f(z) - P(z)| = |-z^8 - z - 2| < |f(z)| = 5$ (Why was this done?)

Hence f and P have the same number of zeros inside the unit circle (How does this follow from the above?) and this number is 3.

 

[Krizalid1]

It's a trick to use Rouché's Theorem, then for $|z|<1$ you can pick "wisely" a term to satisty the conditions.

 

[Dustinsfl]

It's a trick to use Rouché's Theorem, then for $|z|<1$ you can pick "wisely" a term to satisty the conditions.

I need more of an explanation than it is a trick from Rouche's Theorem.

 

[Fantini]

My guess would be that \( |f(z) - P(z)| = |-z^8 -z -2| \leq |z^8| + |z| + |2| = 1+1+2 < |f(z)| = |-5z^3| = 5 \). As for the reasoning of why they have the same number of zeros inside the unit circle I'm still lost as well.

 

[Dustinsfl]

My guess would be that \( |f(z) - P(z)| = |-z^8 -z -2| \leq |z^8| + |z| + |2| = 1+1+2 < |f(z)| = |-5z^3| = 5 \). As for the reasoning of why they have the same number of zeros inside the unit circle I'm still lost as well.

I knew that piece. I wasn't sure why they set up $|f(z) - P(z)|$.

 

[Krizalid1]

You can see the

http://en.wikipedia.org/wiki/Rouché's_theorem#Symmetric_version

of the Rouché's Theorem, and you'll see why the solver does that.

 

[Dustinsfl]

Can someone actually explain this? I have a book and I couldn't figure it out from there so I need something besides a reference.

 

[Krizalid1]

You probably found the answer, but here's a file which may clarify your problems:

http://nathanpfedwards.com/notes/complex/Lecture20110315.pdf