單元 9 Laurent 級數
【例題1】
Classify the singularities of ( ) 1 . ( 1)( 2)
f z = z z
− − Obtain the Laurent expansion centered on z =0 for the regions:
(1) z <1 (2)1< <z 2 (3) z >2 (4)0< − < z 1 1
【參考解答】因為 1
( ) ( 1)( 2)
f z = z z
− − ，所以 z =1、2皆為一階極點。
(1) 當 z <1， 1 3 7 2
( ) 2 4 8
f z = + z+ z +
(2) 當1< <z 2， 1 1 1 2 1 1 12
( ) [1 ] [1 ]
2 2 4
f z z z
z z z
= − + + + − + + +
(3) 當 z >2， 12 32 74 ( )
f z = z + z + z +
(4) 當0< − <z 1 1， 1 2 ( ) 1 ( 1) ( 1)
f z 1 z z
= −z − − − − − −
− ，z =1為一階
極點。
【例題2】
Find the Laurent series of 1 2
( ) (1 )
f z = z z
+ around z0 = for 0 (1) 0< <z 1 (2) 1 z< < ∞ .【91暨南電機】
【參考解答】(1) 1 2 4 6
( ) [1 ]
f z z z z
= z − + − +
(2) 13 15 17 ( )
f z = z − z + z −
【例題3】
Expand ( ) 2 1
( 1)( 2) f z z
z z
= −
+ − into Laurent series centered at z = −1, i.e., into power series in z + 1. You should discuss the expansion in each regions of the complex plane and specify clearly the convergence region of each of your power series.【91清大電機】
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【參考解答】已知 2 1 ( ) ( 1)( 2) f z z
z z
= −
+ − ，
(1)當0< + <z 1 3時， 1 1 12 13 2
( ) ( 1) ( 1)
1 3 3 3
f z z z
= z − − + − + −
+ (2)當3< + < ∞z 1 時，
( ) ( )
2
2 3
2 3 3
( ) 1 1 1
f z = z + z + z +
+ + +
z = −1為一階極點。
【例題4】
Consider f z( ) (= z2−1)−1 where z x iy= + is a complex variable. The Laurent expansion of ( )f z with z = + as the center can be expressed 1 i as f z( )=∑C zn( − −1 i) .n If the region of convergence of this
expansion is 1< − − <z 1 i 5, please find out the coefficients C−2 and C .2 【90中央土木】
【參考解答】 2 1 1 3 2 (2 )
C− = − i
+ ， 2 2
C = −i 。
【例題5】
Let C denote the unit circle z = taken counterclockwise. Show that 1
0
1 1 1
exp( ) (0! 1)
2 C n !( 1)!
z dz
i z n n
π
∞
=
+ = =
∑ +
∫ . 【86清大工科】
【參考解答】
0
1 1 1
exp( )
2 C z dz n !( 1)
i z n n
π
∞
=
+ =
∑ +
∫ ，故得證。
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