Page 59 - Calculus with Complex Numbers
P. 59
Ffgure 5. /
Example 2 Consider the integral
1:: dt
o 1 + sin2 t '
Putting c = eit gives
1:%' dt dz 1 4/c dz
z = z = 4 z ,
0 1 + sin t p. U' 1 - (z - 1/ z ) /4 y z - 6.:: + 1
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where y is the unit circle. Here we used the formula
The singularities of the integrand are at the solutions of the equation
. / - 6z2 + 1 = 0
which are given by
Of these only z = ulu 3 - 2.$,.1' = ulu (x'C - 1) are inside y (Figure 5.2).
/
Differentiating the denominator we obtain the residues by evaluating
4ïz i
4c3 - 12c c2 - 3
at .:2 = 3 - 2./-. Therefore the residues are -i(l.vX at lnoth these points.
2
Hence by the residue theorem we have
