Where the method used should be using complex analysis. int_c d theta/(p+cos theta)^2=2pi p/((p^2-1)sqrt(p^2-1));c:|z|=1

Deon Moran 2022-10-24 Answered
Where the method used should be using complex analysis.
c d θ ( p + cos θ ) 2 = 2 π p ( p 2 1 ) p 2 1 ; c : | z | = 1
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Answers (2)

Kash Osborn
Answered 2022-10-25 Author has 18 answers
i do my self like this on | z | = 1 , z = e i θ , d θ = d z i z
using substitution cos θ = z + z 1 2
1 i d z z ( p + z + z 1 2 ) ( p + z + z 1 2 ) o r 1 i 1 ( 2 p z + z 2 + 1 ) ( p + z + z 1 2 )
and then multiple by z z
1 i z ( 2 p z + z 2 + 1 ) ( 2 p z + z 2 + 1 )
the roots of ( 2 p z + z 2 + 1 ), z 1 = p + p 2 1 ; z 2 = p p 2 1
let z 1 = a and z 2 = b
so a b = 2 p 2 1 and a + b = 2 p
and then i use residue theorm, but the result on the right π p 2 ( p 2 1 ) p 2 1
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Christopher Saunders
Answered 2022-10-26 Author has 6 answers
On | z | = 1, we have z = e i θ , so d z d θ = i e i θ , and hence d θ = d z i z
Also on |z|=1, we have cos θ = 1 2 ( z + 1 z )
Putting this together, you get
1 i | z | = 1 1 z ( p + 1 2 ( z + 1 z ) ) 2   d z .
Expand the denominator,
1 i | z | = 1 1 z ( p + 1 2 ( z + 1 z ) ) 2   d z = 1 i | z | = 1 1 z ( p 2 + p ( z + 1 / z ) + 1 4 ( z + 1 / z ) 2 ) = 1 i | z | = 1 1 ( p 2 z + p z 2 + p + z 4 ( z 2 + 2 + 1 / z 2 ) ) =
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