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[SOLVED] Number of zeros when |z| = 1

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

Krizalid

Active member
Feb 9, 2012
118
It's a trick to use Rouché's Theorem, then for $|z|<1$ you can pick "wisely" a term to satisty the conditions.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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 $|f(z) - P(z)|$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.