Laurent Series

If f(z) is analytic throughout the annular region between and on the concentric circles K_1 and K_2 centered at z=a and of radii r_1 and r_2<r_1 respectively, then there exists a unique series expansion in terms of positive and negative powers of (z-a),

 f(z)=sum_(k=0)^inftya_k(z-a)^k+sum_(k=1)^inftyb_k(z-a)^(-k),
(1)

where

a_k=1/(2pii)∮_(K_1)(f(zeta)dzeta)/((zeta-a)^(k+1))
(2)
b_k=1/(2pii)∮_(K_2)(zeta-a)^(k-1)f(zeta)dzeta
(3)

(Korn and Korn 1968, pp. 197-198).

LaurentSeries

Let there be two circular contours C_2 and C_1, with the radius of C_1 larger than that of C_2. Let z_0 be at the center of C_1 and C_2, and z be between C_1 and C_2. Now create a cut line C_c between C_1 and C_2, and integrate around the path C=C_1+C_c-C_2-C_c, so that the plus and minus contributions of C_c cancel one another, as illustrated above. From the Cauchy integral formula,

f(z)=1/(2pii)int_C(f(z^'))/(z^'-z)dz^'
(4)
=1/(2pii)int_(C_1)(f(z^'))/(z^'-z)dz^'+1/(2pii)int_(C_c)(f(z^'))/(z^'-z)dz^'-1/(2pii)int_(C_2)(f(z^'))/(z^'-z)dz^'-1/(2pii)int_(C_c)(f(z^'))/(z^'-z)dz^'
(5)
=1/(2pii)int_(C_1)(f(z^'))/(z^'-z)dz^'-1/(2pii)int_(C_2)(f(z^'))/(z^'-z)dz^'.
(6)

Now, since contributions from the cut line in opposite directions cancel out,

f(z)=1/(2pii)int_(C_1)(f(z^'))/((z^'-z_0)-(z-z_0))dz^'-1/(2pii)int_(C_2)(f(z^'))/((z^'-z_0)-(z-z_0))dz^'
(7)
=1/(2pii)int_(C_1)(f(z^'))/((z^'-z_0)(1-(z-z_0)/(z^'-z_0)))dz^'-1/(2pii)int_(C_2)(f(z^'))/((z-z_0)((z^'-z_0)/(z-z_0)-1))dz^'
(8)
=1/(2pii)int_(C_1)(f(z^'))/((z^'-z_0)(1-(z-z_0)/(z^'-z_0)))dz^'+1/(2pii)int_(C_2)(f(z^'))/((z-z_0)(1-(z^'-z_0)/(z-z_0)))dz^'.
(9)

For the first integral, |z^'-z_0|>|z-z_0|. For the second, |z^'-z_0|<|z-z_0|. Now use the Taylor series (valid for |t|<1)

 1/(1-t)=sum_(n=0)^inftyt^n
(10)

to obtain

f(z)=1/(2pii)[int_(C_1)(f(z^'))/(z^'-z_0)sum_(n=0)^(infty)((z-z_0)/(z^'-z_0))^ndz^'+int_(C_2)(f(z^'))/(z-z_0)sum_(n=0)^(infty)((z^'-z_0)/(z-z_0))^ndz^']
(11)
=1/(2pii)sum_(n=0)^(infty)(z-z_0)^nint_(C_1)(f(z^'))/((z^'-z_0)^(n+1))dz^'+1/(2pii)sum_(n=0)^(infty)(z-z_0)^(-n-1)int_(C_2)(z^'-z_0)^nf(z^')dz^'
(12)
=1/(2pii)sum_(n=0)^(infty)(z-z_0)^nint_(C_1)(f(z^'))/((z^'-z_0)^(n+1))dz^'+1/(2pii)sum_(n=1)^(infty)(z-z_0)^(-n)int_(C_2)(z^'-z_0)^(n-1)f(z^')dz^',
(13)

where the second term has been re-indexed. Re-indexing again,

 f(z)=1/(2pii)sum_(n=0)^infty(z-z_0)^nint_(C_1)(f(z^'))/((z^'-z_0)^(n+1))dz^' 
 +1/(2pii)sum_(n=-infty)^(-1)(z-z_0)^nint_(C_2)(f(z^'))/((z^'-z_0)^(n+1))dz^'.
(14)

Since the integrands, including the function f(z), are analytic in the annular region defined by C_1 and C_2, the integrals are independent of the path of integration in that region. If we replace paths of integration C_1 and C_2 by a circle C of radius r with r_1<=r<=r_2, then

f(z)=1/(2pii)sum_(n=0)^(infty)(z-z_0)^nint_C(f(z^'))/((z^'-z_0)^(n+1))dz^'+1/(2pii)sum_(n=-infty)^(-1)(z-z_0)^nint_C(f(z^'))/((z^'-z_0)^(n+1))dz^'
(15)
=1/(2pii)sum_(n=-infty)^(infty)(z-z_0)^nint_C(f(z^'))/((z^'-z_0)^(n+1))dz^'
(16)
=sum_(n=-infty)^(infty)a_n(z-z_0)^n.
(17)

Generally, the path of integration can be any path gamma that lies in the annular region and encircles z_0 once in the positive (counterclockwise) direction.

The complex residues a_n are therefore defined by

 a_n=1/(2pii)int_gamma(f(z^'))/((z^'-z_0)^(n+1))dz^'.
(18)

Note that the annular region itself can be expanded by increasing r_1 and decreasing r_2 until singularities of f(z) that lie just outside C_1 or just inside C_2 are reached. If f(z) has no singularities inside C_2, then all the b_k terms in (◇) equal zero and the Laurent series of (◇) reduces to a Taylor series with coefficients a_k.

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