We
now look at more difficult complex integrals:

**÷**z

**÷ < 1**

∫

_{o}^{2}^{p}dq**/**(1 + ½ cos q)
Here we make us of: cos q =
[exp(iq)
+ exp (-iq)]/ 2

But
z = exp (iq) so that:
cos q = z +
z

^{-1}/ z
And
dz = iexp(i q ) dq =
iz dq so that: dq =
dz/ iz

We can
then rewrite the integral in terms of z:

∫

_{o}^{2}^{p}dq**/**(1 + ½ cos q) =**ò**_{C}^{ }dz/ iz / 1 + ½ (z + z^{-1}/ z) =
1/2i

**ò**_{C}^{ }2 dz/ (2z + ½ z^{2}+ ½)
First,
obtain the poles from the quadratic in the denominator:

½ z

^{2}+ 2z + ½ = 0
Solve for z
using the

z = [-b

*quadratic formula*:z = [-b

__+__{b^{2 }- 4ac}^{1/2}]/ 2a
To
obtain:

z=
-2

__+__Ö3 or z1= -2 + Ö3 and z2= -2 - Ö3
We
can eliminate the 2

^{nd}root (pole) since we require:**÷**z**÷ < 1**

**Then:**we use

**-2 + Ö3**

Because
this is a simple pole we can use:

Res
[z]

_{z }_{® }**= p(z)/ q’(z)**_{z o}

Where
q’(z) = z + 2 and p(z) = 4

Then: Res [z]

_{z }_{® }**= 4/ (z + 2)**_{z o}
Insert
z = -2 + Ö3 :

Res
[z]

_{z }_{® }_{-2 + }_{Ö}_{3}= 4/ [(-2 + Ö3) + 2] = 4/Ö3
Then:

**ò**

_{C}

^{ }f(z) dz = (1/ 2i) [2 pi c

**] = 4p / Ö3**

_{- 1}__Problems for Math Mavens__:

1)
Evaluate:

∫

_{o}^{2}^{p}dq**/**(37 - 12 cos q) for**÷**z**÷**__<__6
2)Evaluate:

∫

_{o}^{2}^{p}dq**/**(13 - 5 sin q) for**÷**z**÷**__<__i
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