will be of great assistance in dealing with Laurent series.
Example 12-21. (Laurent series)
Express the function f(z) = z
(z−1)(z−3)as a Laurent series centered at the
singular point z= 1.
Solution
The problem is to express f(z)as a series involving both negative and positive
powers of (z−1).To accomplish this task analyze the following two representations
of f(z)
(i) f(z) = [(z−1) + 1]
(z−1)[(z−1) −2]=[
1 +^1
z− 1]
[(z−1) −2]−^1(ii) f(z) =[(z−1) + 1]
(z−1)[(z−1) −2] =−[
1 +1
z− 1]
[2 −(z−1)]−^1The last term in the representations (i) and (ii) have binomial expansions having
the form of equation (12.195). The binomial expansion for the last term in repre-
sentation (i) converges for 2 <|z− 1 |and the binomial expansion for the last term in
representation (ii) converges for |z− 1 |< 2 .The term (z−1) > 0 is assumed to hold in
both the representations (i) and (ii). Therefore, to get the annular region isolating
the singular point at z= 1 the representation (ii) is used. Expanding the last term
in the representation (ii) gives
f(z) = −[
1 +^1
z− 1][
1
2+^1
22(z−1) +^1
23(z−1)^2 +^1
24(z−1)^3 +···+^1
2 n+1(z−1)n+···](a) Multiply the first term in the representation (ii) by the expanded second
term and then collect like terms to obtain the Laurent series expansion
f(z) = −^1 /^2
z− 1−^3
22−^3
23(z−1) −^3
24(z−1)^2 −^3
25(z−1)^3 −^3
26(z−1)^4 −··· (12.196)