# ? f s s zds 12 πi z γ 2 f s s z ds z γ1 f s s z ds

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η f ( s ) s - z ds ! = 1 2 π i Z γ 2 f ( s ) s - z ds - Z γ 1 f ( s ) s - z ds ! = 1 2 π i Z γ 2 f ( s ) s - z ds + Z γ 1 f ( s ) z - s ds ! . We have 1 s - z = 1 s 1 - z s = 1 s X z s n whenever | z s | < 1. Similarly, 1 z - s = 1 z X s z n ,
2. PROOF OF THEOREM ?? 65 whenever | s z | < 1. Substituting these two expressions into the above formula we get f ( z ) = 1 2 π i Z γ 2 X f ( s ) 1 s z s n ! ds + Z γ 1 X f ( s ) 1 z s z n ! ds ! We can exchange the integration and summation, since the series converges uniformly, to obtain f ( z ) = 1 2 π i X n = 0 Z γ 2 f ( s ) s n + 1 z n ds ! + 1 2 π i - 1 X n = -∞ Z γ 1 f ( s ) s n + 1 z n ds ! . Now, by the Cauchy-Goursat Theorem for multiply connected domains (Theorem 5.16) we have for each n Z γ 1 f ( s ) s n + 1 z n ds = Z S (0 , r ) f ( s ) s n + 1 z n ds and Z γ 2 f ( s ) s n + 1 z n ds = Z S (0 , r ) f ( s ) s n + 1 z n ds . Thus f ( z ) = X n = -∞ c n z n where c n = 1 2 π i Z S (0 , r ) f ( s ) s n + 1 ds , as claimed. E xample . Find the Laurent (or Taylor) series expansions about z = 0 for f ( z ) = z 2 ( z - 1)( z 2 + 4) . Now f ( z ) has isolated singularities at z = 1 and z = ± 2 i , so there are three expansions; for (a) | z | < 1; (b) 1 < | z | < 2; (c) | z | > 2. Express f in terms of partial fractions: f ( z ) = - 1 5 1 1 - z - 4 + 4 z 4 + z 2 ! . Consider the first term: For | z | < 1, 1 1 - z = 1 + z + z 2 + . . . = X k = 0 z k . For | z | > 1, 1 - z (1 - 1 / z ) = - 1 z 1 + 1 z + 1 z 2 + . . . ! = - 1 z - 1 z 2 - 1 z 3 - . . . . Consider the second term:
66 6. ROOTS AND SINGULARITIES For | z | < 2, 1 + z 1 + z 2 / 4 = (1 + z ) X k = 0 ( - 1) k z 2 k 4 k = X k = 0 ( - 1) k z 2 k 4 k + X k = 0 ( - 1) k z 2 k + 1 4 k . For | z | > 2, 4 + 4 z 4 + z 2 = 4 z 2 1 + z 1 + 4 / z 2 = 4 z 2 (1 + z ) X k = 0 ( - 1) k 4 k z - 2 k = X k = 0 ( - 1) k 4 k z - 2 k - 2 + X k = 0 ( - 1) k 4 k z - 2 k - 1 Then we have (a) for | z | < 1 f ( z ) = - 1 5 1 + z + z 2 + z 3 + z 4 - 1 + z - z 2 4 - z 3 4 + z 4 16 + . . . !! = - z 2 4 - z 3 4 - 3 z 4 16 - . . . . (b) for 1 < | z | < 2 f ( z ) = - 1 5 - 1 z - 1 z 2 - 1 z 3 - . . . - 1 + z - z 2 4 - z 3 4 + . . . !! = 1 5 1 z + 1 z 2 + 1 z 3 + . . . ! + 1 5 1 + z - z 2 4 - z 3 4 + . . . ! . (c) for | z | > 2 f ( z ) = - 1 5 - 1 z - 1 z 2 - 1 z 3 - . . . - 4 z + 4 z 2 - 16 z 3 - . . . !! = 1 z + 1 z 2 - 3 z 3 - . . . . 3. Zeros 3.1. Zeros and the Unique Continuation Theorem. D efinition 6.6. If f H ( Ω ), then any point a such that f ( z 0 ) = 0 is called a zero of f . The zero z 0 is said to be isolated if f ( z 0 ) = 0, and there exists a δ > 0 such that f ( z ) , 0 for all z D 0 ( z 0 , δ ).
3. ZEROS 67 An isolated zero z 0 is said to be of order m 1 if the Taylor series has the form f ( z ) = X n m a n ( z - z 0 ) n , a m , 0 , i.e. a 0 = a 1 = · · · = a m - 1 = 0. Notation for the set of zeroes: Z ( f ). T heorem 6.7. Let f H ( Ω ) with a domain Ω . Suppose that the set of zeroes Z ( f ) has an accumulation point in Ω . Then f = 0 in Ω . P roof . Let z 0 Ω be the accumulation point of the set Z ( f ). Let r > 0 be a number such that D ( z 0 , r ) Ω . By Taylor’s Theorem, f ( z ) = X n 0 a n ( z - z 0 ) n in D ( z 0 , r ). Assume that f ( z ) is not identically zero in D ( z 0 , r ). Then there is at least one coe ffi cient a n , 0. Let a m , m 1 , be the first non-zero coe ffi cient. Thus f ( z ) = ( z - z 0 ) m X n m a n ( z - z 0 ) n - m = ( z - z 0 ) m ( a m + a m + 1 z + . . . ) . This power series in the brackets has the same radius of convergence as the series for f , and hence it defines a continuous function. This function tends to a m as z z 0 , so that there is a disk D ( z 0 , ) such that a m + a m + 1 z + · · · , 0 for all z D ( z 0 , ). Thus the function f has no zeroes in D ( z 0 , ) except for z 0 . This means that Z ( f ) cannot have z 0 as an accumulation point.
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