# Complex Variables: Missing Step in Example

In summary, the integral from 0 to 2*pi of [(Cos(3*theta)) / (5 - 4Cos(theta))] (d*theta) can be solved by factoring out z^{-3} and using the 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, and dz = (i*z) (d*theta). By factoring the quadratic 2z^2-5z+2, the integral can be simplified to -(1

## 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!

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!