Singularities/Residues/ Applications
9.8 Some Special Rules for the Computation of Residues
669(a) Residue at a finite simple pole. If a i= oo is a simple pole of f(z), we
have the Laurent representation
f(z)= A_^1 +Ao+A1(z-a)+···
z-a
valid in 0 < lz - al < 8 for some 8 > 0. Hence
g(z) = (z - a)f(z) = A_ 1 + Ao(z - a)+···
is regular at a, and we have
g(a) = lim(z - a)f(z) = A-1 = Resf(z)z-a z=a
Example To find Resz=-n r(z). Since
r(z) - r( z + n + 1)
- z(z+l)···(z+n-l)(z+n)
we obtain
r(l) (-l)n
lim (z + n)r(z) = = -
z--n -n(-n + 1) · · · (-1) n!In particular, suppose that f(z) is given in the form
f( ) = g(z)
z h(z)(9.8-1)where both g and h are regular at a and g(a) i= 0, h(a) = 0, h'(a) i= 0.
Clearly, in this case a is a simple pole of f(z), and we have
g(z) 1/J(z)
f(z) = (z - a)h 1 (z) = z -a
where 1/J(z) = g(z)/h 1 (z) is regular at a, and 1/J(a) = g(a)/h 1 (a) i= 0.
Then, by applying (9.8-1) we get
. g(a) g(a)
Resf(z) z=a = hm(z z-a - a)f(z) = -h 1 ( ) a = h'( a ) (9.8-2)
since h(z) = (z - a)h1(z) implies that h'(a) = h1(a).
Example Let f(z) = tanz = sinz/ cosz. We have sin 1/ 2 7r = 1, cos%7r =
O, and (cosz)' = -1 for z = %7r. Hence^1 / 2 7r is a simple pole of tanz, and
sin^1 / 2 7r -
Res tanz =. 1 / = -1
z=l/2n: - Sln 21r