how many internal zeros does a polynomial has 3z^2 + 8z^2 - 1
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Answer:
to use Rouche's theorem.
So I tried to find a function g that has 2 zeros in the unit disk and:
|f(z)−g(z)|<|f(z)|+|g(z)|∀z∈D(1)
However, I couldn't find such function.
I tried g(z)=3z4+9z3+10. This function has 2 zeros in the unit disk according to Wolfram Alpha. I wasn't able to prove (1) and that g has 2 zeros in the unit disk with an analytical method.
I did the same for the function g(z)=z5+9z3+10 that has two zeros in unit disk by Wolfram Alpha. It didn't work either.
Could somebody help out to prove that f has 2 zeros in the unit disk?
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Jun 15 '17 at 17:51
user344460
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tough problem. Working on it using g(z)=(z+3/2+8i/3)(z+3/2−8i/3)(z+9/8)(z−9/16+7i/9)(z−9/16+7i/9) which seems to work but its hard to show it. – Mark Fischler Jun 15 '17 at 19:06
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Correct me if I'm wrong, but given |z|=1, then z¯=1z, and z≠0. So if z5+3z4+9z3+10=0, then 1z5+3z4+9z3+10=0, so z is also a root of 1+3z+9z2+10z5. It is then a root of the gcd of those two polynomials, but wolfram says it's gcd is 1, so someone is wrong here. – Henrique Augusto Souza Jun 15 '17 at 19:21
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My proposed g(z) fails the condition by a tiny amount at z=2930e3π10. – Mark Fischler Jun 15 '17 at 19:38
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This is a near duplicate. See math.stackexchange.com/q/2038467/254075 – sharding4 Jun 15 '17 at 20:10
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@HenriqueAugustoSouza Yes, the unit disc is the open domain |z|<1; it's boundary is the unit circle |z|=1. That f doesn't vanish along the unit circle is at the basis of the attempts to use Rouche's theorem, here. – Jonathan Y. Jun 15 '17 at 22:50
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This is a technique to solve the more general problem of counting the number of zeros of a polynomial inside the unit circle. One could use it for other curves