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6. Residue Calculus Let Z0be an Isolated Singularity of F(Z)

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6. Residue Calculus Let Z0be an Isolated Singularity of F(Z) 6. Residue calculus Let z0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood Nε = z : 0 < z z < ε such that f is { | − 0| } analytic everywhere inside Nε. We define 1 Res(f,z0)= f(z) dz, 2πi IC where C is any simple closed contour around z0 and inside Nε. 1 Since f(z) admits a Laurent expansion inside Nε, where ∞ n ∞ n f(z)= an(z z ) + bn(z z ) , − 0 − 0 − nX=0 nX=1 then 1 b1 = f(z) dz = Res(f,z0). 2πi IC Example 1 1 if k = 1 Res ,z0 = (z z )k ! ( 0 if k = 1 − 0 6 1 1 2 1 Res(e1/z, 0) = 1 since e1/z =1+ + + , z > 0 1! z 2!z2 ··· | | 1 1 Res , 1 = by the Cauchy integral formula. (z 1)(z 2) ! 1 2 − − − 2 Cauchy residue theorem Let C be a simple closed contour inside which f(z) is analytic ev- erywhere except at the isolated singularities z ,z , ,zn. 1 2 ··· f(z) dz = 2πi[Res (f,z1)+ + Res(f,zn)]. IC ··· This is a direct consequence of the Cauchy-Goursat Theorem. 3 Example Evaluate the integral z + 1 dz z =1 z2 I| | using (i) direct contour integration, (ii) the calculus of residues, 1 (iii) the primitive function log z . − z Solution (i) On the unit circle, z = eiθ and dz = ieiθ dθ. We then have 2π 2π z + 1 iθ 2iθ iθ iθ dz = (e− +e− )ie dθ = i (1+e− ) dθ = 2πi. z =1 z2 0 0 I| | Z Z 4 (ii) The integrand (z + 1)/z2 has a double pole at z = 0. The Laurent expansion in a deleted neighborhood of z = 0 is simply 1 1 + , where the coefficient of 1/z is seen to be 1. We have z z2 z + 1 Res 0 = 1 2 , , z and so z + 1 z + 1 dz = 2πiRes , 0 = 2πi. z =1 z2 z2 I| | (iii) When a closed contour moves around the origin (which is the branch point of the function log z) in the anticlockwise direction, the increase in the value of arg z equals 2π. Therefore, z + 1 = change in value of ln + arg 1 in 2 dz z i z z I z =1 z | | − | | traversing one complete loop around the origin = 2πi. 5 Computational formula Let z0 be a pole of order k. In a deleted neighborhood of z0, b b f(z)= ∞ a (z z )n + 1 + + k , b = 0. n 0 k k − z z0 ··· (z z0) 6 nX=0 − − Consider g(z) = (z z )kf(z). − 0 the principal part of g(z) vanishes since k 1 ∞ n+k g(z)= bk + bk 1(z z0)+ b1(z z0) − + an(z z0) . − − ··· − − nX=0 By differenting (k 1) times, we obtain − (k 1) g − (z0) if (k 1)( ) is analytic at (k 1)! g − z z0 − k 1 k d (z z0) f(z) b1 = Res(f,z0)= lim − − if z is a removable z z k 1 0 → 0dz − " (k 1)! # − (k 1) singularity of g − (z) 6 Simple pole k =1: Res(f,z )= lim (z z )f(z). 0 z z 0 → 0 − p(z) Suppose f(z)= where p(z0) = 0 but q(z0) = 0, q (z0) = 0. q(z) 6 ′ 6 Res(f,z ) = lim (z z )f(z) 0 z z 0 → 0 − p(z0)+ p′(z0)(z z0)+ = lim (z z0) − ··· z z0 q (z0) 2 → − q (z )(z z )+ ′′ (z z ) + ′ 0 − 0 2! − 0 ··· p(z ) = 0 . q′(z0) 7 Example Find the residue of e1/z f(z)= 1 z − at all isolated singularities. Solution (i) There is a simple pole at z = 1. Obviously Res(f, 1) = lim (z 1)f(z)= e1/z = e. z 1 − − − → z=1 (ii) Since 1 1 e1/z =1+ + + z 2!z2 ··· has an essential singularity at z = 0, so does f(z). Consider e1/z 1 1 = (1+z +z2 + ) 1+ + + , for 0 < z < 1, 1 z ··· z 2!z2 ··· | | − the coefficient of 1/z is seen to be 1 1 1+ + + = e 1 = Res(f, 0). 2! 3! ··· − 8 Example Find the residue of z1/2 f(z)= z(z 2)2 − at all poles. Use the principal branch of the square root function z1/2. Solution The point z = 0 is not a simple pole since z1/2 has a branch point at this value of z and this in turn causes f(z) to have a branch point there. A branch point is not an isolated singularity. However, f(z) has a pole of order 2 at z = 2. Note that d z1/2 z1/2 1 Res(f, 2) = lim = lim = , z 2 dz z ! z 2 − 2z2 ! −4√2 → → where the principal branch of 21/2 has been chosen (which is √2). 9 Example Evaluate Res(g(z)f ′(z)/f(z),α) if α is a pole of order n of f(z), g(z) is analytic at α and g(α) = 0. 6 Solution Since α is a pole of order n of f(z), there exists a deleted neigh- borhood z : 0 < z α < ε such that f(z) admits the Laurent { | − | } expansion: bn b b f(z)= + n 1 + + 1 + ∞ a (z α)n, b = 0. n − n 1 n n (z α) (z α) − ··· (z α) − 6 − − − nX=0 Within the annulus of convergence, we can perform termwise dif- ferentiation of the above series 10 nbn (n 1)bn b1 ∞ n 1 f ′(z)= − − + nan(z α) − . (z α)n+1 − (z α)n −···− (z α)2 − − − − nX=0 Provided that g(α) = 0, it is seen that 6 (z α) nbn (n 1)bn b1 + na (z α)n 1 (z −α)n+1 (z− α)n (z α)2 n∞=0 n − = lim ( ) − − − − −···− − − g z b z α bn + n 1 + + b1 + P a (z α)n → (z α)n (z α−)n 1 z α n∞=0 n − − − ··· − − = ng(α) = 0, P − 6 so that α is a simple pole of g(z)f ′(z)/f(z). Furthermore, f Res g ′,α = ng(α). f ! − Remark When g(α) = 0,α becomes a removable singularity of gf ′/f. 11 Example Suppose an even function f(z) has a pole of order n at α. Within the deleted neighborhood z : 0 < z α < ε , f(z) admits the Laurent { | − | } expansion bn b1 ∞ n f(z)= + + + an(z α) , bn = 0. (z α)n ··· (z α) − 6 − − nX=0 Since f(z) is even, f(z)= f( z) so that − bn b1 ∞ n f(z)= f( z)= + + + an( z α) , − ( z α)n ··· ( z α) − − − − − − nX=0 which is valid within the deleted neighborhood z : 0 < z + α < ε . { | | } Hence, α is a pole of order n of f( z). Note that − − Res(f(z),α)= b and Res(f(z), α)= b 1 − − 1 so that Res(f(z),α) = Res(f(z), α). For an even function, if − − z = 0 happens to be a pole, then Res(f, 0) = 0. 12 Example tan z tan z π tan z π dz = 2πi Res , + Res , z =2 z z 2 z −2 I| | π since the singularity at z = 0 is removable. Observe that is a 2 π simple pole and cos z = sin z , we have − − 2 π tan z π z 2 tan z Res , = lim − z 2 z π z →2 z π sin z = lim − 2 π π 3 z 2 π (z 2) → z z 2 + −6 + "− − ··· # 1 2 = = . π −π −2 13 tan z π 2 As tan z/z is even, we deduce that Res , = using the z −2 π result from the previous example. We then have tan z dz = 0. z =2 z I| | Remark π 2 π Let p(z) = sin z/z,q(z) = cos z, and observe that p = , q = 2 π 2 π 0 and q′ = 1 = 0, then 2 − 6 tan z π π π 2 Res , = p q′ = − . z 2 2 , 2 π 14 Example Evaluate z2 2 2 2 dz. IC (z + π ) sin z Solution z z z z lim = lim lim = 0 z 0 sin z (z2 + π2)2 z 0 sin z z 0 (z2 + π2)2! → → → so that z = 0 is a removable singularity. 15 It is easily seen that z = iπ is a pole of order 2. d Res(f,iπ) = lim [(z iπ)2f(z)] z iπ dz − → d z2 = lim z iπ dz "(z + iπ)2 sin z# → 2z(z + iπ) sin z z2[(z + iπ) cos z + 2 sin z] = lim − z iπ (z + iπ)3 sin2 z → 2 sinh π + ( π cosh π sinh π) 1 cosh π = − − = + . 4π sinh2 π −4π sinh π 4π sinh2 π − Recall that sin iπ = i sinh π and cos iπ = cosh π. Hence, z2 = 2 Res( ) 2 2 2 dz πi f,iπ IC (z + π ) sin z i 1 cosh π = + 2 . 2 −sinh π sinh π 16 Theorem If a function f is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed contour C, then 1 1 ( ) = 2 Res 0 f z dz πi 2f , . IC z z 17 We construct a circle z = R which is large enough so that C is | | 1 interior to it. If C denotes a positively oriented circle z = R , 0 | | 0 where R0 > R1, then ∞ n f(z)= cnz , R1 < z < , (A) n= | | ∞ X−∞ where 1 f(z) c = dz n = 0, 1, 2, .
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