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For z 0 , define f ( z ) = e 1 / z . Show that 0 is an essential singularity of f .

Let f be continuous on a punctured disk B ' ¯ r ( c ) , analytic at each point of B r ' ( c ) , and suppose that c is an essential singularity of f . Then

  1. For all z B r ' ( c ) ,
    f ( z ) = k = - a k ( z - c ) k ,
    where the sequence { a k } - has the property that for any negative integer N there is a k < N such that a k 0 .
  2. The infinite series in part (1) converges uniformly on each compact subset K of B r ' ( c ) . That is, if F n is defined by F n ( z ) = k = - n n a k ( z - c ) k , then the sequence { F n } converges uniformly to f on the compact set K .
  3. For any piecewise smooth geometric set S B r ( c ) , whose boundary C S has finite length, and satisfying c S 0 , we have
    C S f ( ζ ) d ζ = 2 π i a - 1 ,
    where a - 1 is the coefficient of ( z - c ) - 1 in the series of part (1).

Define numbers { a k } - as follows.

a k = 1 2 π i C r f ( ζ ) ( ζ - c ) k + 1 d ζ .

Note that for any 0 < δ < r we have from Cauchy's Theorem that

a k = 1 2 π i C δ f ( ζ ) ( ζ - c ) k + 1 d ζ ,

where C δ denotes the boundary of the disk B ¯ δ ( c ) .

Let z c be in B r ( c ) , and choose δ > 0 such that δ < | z - c | . Then, using part (c) of [link] , and then mimicking the proof of [link] , we have

f ( z ) = 1 2 π i C r f ( ζ ) ζ - z d ζ - 1 2 π i C δ f ( ζ ) ζ - z d ζ = 1 2 π i C r f ( ζ ) ( ζ - c ) - ( z - c ) d ζ + 1 2 π i C δ f ( ζ ) ( z - c ) - ( ζ - c ) d ζ = 1 2 π i C r f ( ζ ) ζ - c 1 1 - z - c ζ - c d ζ + 1 2 π i C δ f ( ζ ) z - c 1 1 - ζ - c z - c d ζ = 1 2 π i C r f ( ζ ) ζ - c k = 0 ( z - c ζ - c ) k d ζ + 1 2 π i C δ f ( ζ ) z - c j = 0 ( ζ - c z - c ) j d ζ = k = 0 1 2 π i C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k + j = 0 1 2 π i C δ f ( ζ ) ( ζ - c ) j d ζ ( z - c ) - j - 1 = k = 0 a k ( z - c ) k + k = - - 1 1 2 π i C δ f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = k = 0 a k ( z - c ) k + k = - - 1 1 2 π i C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = k = - a k ( z - c ) k ,

which proves part (1).

We leave the proofs of parts (2) and (3) to the exercises.

  1. Justify bringing the summation signs out of the integrals in the calculation in the preceding proof.
  2. Prove parts (2) and (3) of the preceding theorem. Compare this with [link] .

REMARK The representation of f ( z ) in the punctured disk B r ' ( c ) given in part (1) of [link] and [link] is called the Laurent expansion of f around the singularity c . Of course it differs from a Taylor series representation of f , as this one contains negative powers of z - c . In fact, which negative powers it contains indicates what kind of singularity the point c is.

Non removable isolated singularities of a function f share the property that the integral of f around a disk centered at the singularity equals 2 π i a - 1 , where the number a - 1 is the coefficient of ( z - c ) - 1 in the Laurent expansion of f around c . This number 2 π i a - 1 is obviously significant, and we call it the residue of f at c, and denote it by R f ( c ) .

Combining [link] , [link] , and [link] , we obtain:

Residue theorem

Let S be a piecewise smooth geometric set whose boundary has finite length, let c 1 , ... , c n be points in S 0 , and suppose f is a complex-valued function that is continuous at every point z in S except the c k 's, and differentiable at every point z S 0 except at the c k 's. Assume finally that each c k is a nonremovable isolated singularity of f . Then

C S f ( ζ ) d ζ = k = 1 n R f ( c k ) .

That is, the contour integral around C S is just the sum of the residues inside S .

Prove [link] .

Use the Residue Theorem to compute C S f ( ζ ) d ζ for the functions f and geometric sets S given below. That is, determine the poles of f inside S , their orders, the corresponding residues, and then evaluate the integrals.

  1. f ( z ) = sin ( 3 z ) / z 2 , and S = B ¯ 1 ( 0 ) .
  2. f ( z ) = e 1 / z , and S = B ¯ 1 ( 0 ) .
  3. f ( z ) = e 1 / z 2 , and S = B ¯ 1 ( 0 ) .
  4. f ( z ) = ( 1 / z ( z - 1 ) ) , and S = B ¯ 2 ( 0 ) .
  5. f ( z ) = ( ( 1 - z 2 ) / z ( 1 + z 2 ) ( 2 z + 1 ) 2 ) , and S = B ¯ 2 ( 0 ) .
  6. f ( z ) = 1 / ( 1 + z 4 ) = ( 1 / ( z 2 - i ) ( z 2 + i ) ) , and S = B ¯ r ( 0 ) for any r > 1 .

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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