1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

8. 1 • THE RESIDUE THEOREM 297

If z 0 is any one of the singularities of f, then we can use L'Hopital 's rule to

compute Reslf,zo]:

R [f I^1

. z - zo l' 1 1
es zo = 1m --= 1m - = - ..
' •-•o z4 + 4 z-•o 4z^3 4z3

Since z~ = -4, we can simplify th.is expression further to yield Res[f, zo]

• 116 z 0 • Hence Res[f, 1 + i] = -~6;, and Res[f , 1 - i) = -:r. We now use the
  resid ue theorem to get

J

--dz =2ni ·(-1-i --+---l+i) =-. -ni
z^4 + 4 16 16 4
Ci°(l)

The theory of residues can be used to expand the quotient of two polynomials

into its partial fraction representation.

• EXAMPLE 8.7 Let P (z) be a polynomial of degree at most 2. Show that
  if a, b, and c are distinct complex numbers, then

f (z) = P(z) = ~ + ~ + __£__'

(z-a)(z-b)(z-c) z - a z-b z-c

where

P(a)

A =Res(/,a] = (a- b)(a- c)'

P(b)

B = Res If, b) = (b - a)(b -c)'

P(c)

C = Res(f,c] = (c-a) (c-b)'

and

So lu tion It will suffice to prove that A = Res[f, a]. We expand f in its Laurent

series about the point a by writing the three terms .~,,. z~b, and zc:._c in their

Laurent series about the point a and adding them. The term "~" is itself a one-

term Laurent series about the point a. The term z~b is analytic at the point a,

and its Laurent series is actually a Taylor series given by

B - B 1

00

B n

z - b = (b -a) (1 - ~) = -~ (b - at+l (z - a) '

which is valid for lz -a.I < lb -al. Likewise, the expansion of the term .C:..c is

C

00

C n

z - c = - I: (c - ar+l (z -a) ,

n=O