1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

296 CHAPTER 8 • RESIDUE THEORY

This last limit involves an indeterminate form, which we evaluate by using

L'Hopital's rule:

Res [f, OJ = 7r 2 Jim - ?r

2

z sin (7rz) + 7r cos (7rz) - 7r cos ( 7rZ)

z-+O 37rSin^2 (7rz)cos(u)

= 1r^2 1· !ID- ------?rZ

Z·-+0 3sin (7rz) cos (7rz)

-7r 2 1· 1rZ Ji 1 - 7r2

= --JID ID =

3 z-+0 sin (7rz) z-+0 cos (7rz) 3

•EXAMPLE 8.5 Find f z•+zL 2 ,1dz.

ct(O)

Solution We write the integrand as f (z) = •"(z+faz-l). The singularities of

f that lie inside C3 (0) are simple poles at the points 1 and - 2, and a pole of

order 2 a t the origin. We compute the residues as follows:

d - 2z - 1 - 1

Res [f,OJ = Jim -d [z^2 f (z)] = lim 2 = -

4

,

z- o z z-O (z2 + z - 2)

1 1

Res [f, l] = l~ (z - 1) f (z) = .f..el z 2 (z + 2 ) = 3' and

Res [f , -2] = lim (z + 2) f (z) = lim 2 (

1

) = -

2

1

.

Z-+-2 Z-+-2 Z Z - 1 1

Finally, the residue theorem yields

J

__ d_z __ = 27ri [--1 + ~ - 2.) = 0
z^4 + z3 - 2z2 4 3 12 ·
ct(o)

The answer, J z•+•1'_ 2 z> = 0, is not at all obvious, and all the preceding

ct(O)

calculations are required to get it.

• EXAMPLE 8.6 F ind f (z^4 + 4)-

1

dz.

ct(1J

Solution The singularities of the integrand f (z) = -:f-that lie inside C2 (1)

z +4

are simple poles occurring at the points 1 ± i , as the points -1 ± i lie outside

C2 (1). Factoring the denominator is tedious, so we use a different approach.