Complex Variables: Missing Step in Example

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SUMMARY

The discussion centers on solving the integral from 0 to 2π of [(Cos(3*theta)) / (5 - 4Cos(theta))] d*theta using complex variables. Key transformations include substituting z = exp(i*theta) and expressing Cos(3*theta) in terms of z, leading to the integral being rewritten as -(1/2i)∫[(z^(6) + 1) / (z^(3)(2z - 1)(z - 2))] dz. The participants clarify the steps to factor the denominator and manipulate the integral for easier computation, emphasizing the importance of proper substitution and factoring techniques in complex analysis.

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Students and professionals in mathematics, particularly those studying complex analysis, as well as anyone looking to enhance their skills in evaluating integrals involving trigonometric functions using complex variables.

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Homework Statement


Integral from 0 to 2*pi of [(Cos(3*theta)) / (5 - 4Cos(theta))] (d*theta)


Homework Equations


z=exp(i*theta),
Cos(theta) = (z + z^(-1)) / 2,
Cos(3*theta) = ((exp(3*i*theta)+exp(-3*i*theta)) / 2 = (z^(3) + z^(-3)) / 2,
dz = (i*z) (d*theta)


The Attempt at a Solution


This was an example in my book that didnt show all the steps T_T
Heres what they did:
=>(integral of...){[(z^(3) + z^(-3))/2] / [5 - 4(z + z^(-1))} (dz/iz)
==>-(1/2i)(integral sign)[(z^(6) + 1) / [(z^(3))(2z - 1)(z - 2)]] (dz)

i don't know what voodoo magic they pulled there, but i would like to find out!
thanks for your time!
 
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Factor out z^{-3} on the top, pull the 2/i right out of the integral, multiply the denominator through with that z that was dividing dz. You are now almost there...

Multiply the denominator by -1, and put a -1 outside of the integral so that you haven't changed the expression. The denominator should now look like z^3(2z^2-5z+2) now you must factor that quadratic 2z^2-5z+2 = (2z - 1)(z - 2).
 
Huzzah! i can see clearly now(!)...
thanks a bunch! now (hopefully) i can apply that stuff in my homework problems!
 

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