Upcoming SlideShare
Loading in...5
×

Like this presentation? Why not share!

# Complex varible

## on Mar 31, 2013

• 341 views

### Views

Total Views
341
Views on SlideShare
341
Embed Views
0

Likes
0
Downloads
1
Comments
0

No embeds

### Upload Details

Uploaded via as Microsoft PowerPoint

### Usage Rights

© All Rights Reserved

### Report content

• Comment goes here.
Are you sure you want to
Your message goes here
Edit your comment

## Complex variblePresentation Transcript

• f ( z ) = u( x , y ) + iv ( x , y ) for z = x + iy f ( z + ∆z ) − f ( z ) f ( z ) = lim [ ] exists ∆z →0 ∆zIts value does not depend on the direction.Ex : Show that the function f ( z ) = x 2 − y 2 + i 2 xy is differentiable for all values of z .for ∆z = ∆x + i∆y f ( z + ∆z ) − f ( z ) f ( z ) = lim ∆z →0 ∆z ( x + ∆x ) 2 − ( y + ∆y ) 2 + 2i ( x + ∆x )( y + ∆y ) − x 2 + y 2 − 2ixy = ∆x + i∆y ( ∆x ) 2 − ( ∆y )2 + 2i∆x∆y = 2x + i2 y + ∆x + i∆y(1) choose ∆y = 0, ∆x → 0 ⇒ f ( z ) = 2 x + i 2 y(2) choose ∆x = 0, ∆y → 0 ⇒ f ( z ) = 2 x + i 2 y
• * * Another method : f ( z ) = ( x + iy ) 2 = z 2 ( z + ∆z )2 − z 2 ( ∆z )2 + 2 z∆z f ( z ) = lim [ ] = lim [ ] ∆z →0 ∆z ∆z → 0 ∆z = lim ∆z + 2 z = 2 z ∆z → 0Ex : Show that the function f ( z ) = 2 y + ix is not differentiable anywhere in the complex plane.f ( z + ∆z ) − f ( z ) 2 y + 2∆y + ix + i∆x − 2 y − ix 2∆y + i∆x = = ∆z ∆x + i∆y ∆x + i∆yif ∆z → 0 along a line thriugh z of slope m ⇒ ∆y = m ∆x f ( z + ∆z ) − f ( z ) 2 ∆ y + i∆ x 2m + i f ( z ) = lim = lim [ ]= ∆z →0 ∆z ∆x ,∆y →0 ∆x + i∆y 1 + imThe limit depends on m (the direction), so f ( z )is nowhere differentiable.
• Ex : Show that the function f ( z ) = 1 /(1 − z ) is analytic everywhere except at z = 1. f ( z + ∆z ) − f ( z ) 1 1 1f ( z ) = lim [ ] = lim [ ( − )] ∆z →0 ∆z ∆z →0 ∆z 1 − z − ∆z 1 − z 1 1 = lim [ ]= ∆z →0 (1 − z − ∆z )(1 − z ) (1 − z ) 2Provided z ≠ 1, f ( z ) is analytic everywhere such thatf ( z ) is independent of the direction.
• 20.2 Cauchy-Riemann relation A function f(z)=u(x,y)+iv(x,y) is differentiable and analytic, there must be particular connection between u(x,y) and v(x,y) f ( z + ∆z ) − f ( z )L = lim [ ] ∆z →0 ∆z f ( z ) = u( x , y ) + iv ( x , y ) ∆z = ∆x + i∆y f ( z + ∆z ) = u( x + ∆x , y + ∆y ) + iv ( x + ∆x , y + ∆y ) u( x + ∆x , y + ∆y ) + iv ( x + ∆x , y + ∆y ) − u( x , y ) − iv ( x , y )⇒ L = lim [ ] ∆x ,∆y →0 ∆x + i∆y(1) if suppose ∆z is real ⇒ ∆y = 0 u( x + ∆x , y ) − u( x , y ) v ( x + ∆x , y ) − v ( x , y ) ∂u ∂v⇒ L = lim [ +i ]= +i ∆x →0 ∆x ∆x ∂x ∂x(2) if suppose ∆z is imaginary ⇒ ∆x = 0 u( x , y + ∆y ) − u( x , y ) v ( x , y + ∆y ) − v ( x , y ) ∂u ∂v⇒ L = lim [ +i ] = −i + ∆y →0 i ∆y i ∆y ∂y ∂y ∂u ∂v ∂v ∂u = and =- Cauchy - Riemann relations ∂x ∂y ∂x ∂y
• Ex : In which domain of the complex plane is f ( z ) =| x | − i | y | an analytic function?u( x , y ) =| x |, v ( x , y ) = − | y | ∂u ∂v ∂ ∂(1) = ⇒ | x |= [− | y |] ⇒ (a) x > 0, y < 0 the fouth quatrant ∂x ∂y ∂x ∂y (b) x < 0, y > 0 the second quatrant ∂v ∂u ∂ ∂(2) =− ⇒ [− | y |] = − | x | ∂x ∂y ∂x ∂yz = x + iy and complex conjugate of z is z * = x − iy⇒ x = ( z + z * ) / 2 and y = ( z − z * ) / 2i ∂f ∂f ∂x ∂f ∂y 1 ∂u ∂v i ∂v ∂u⇒ = + = ( − )+ ( + ) ∂z * ∂x ∂z * ∂y ∂z * 2 ∂x ∂y 2 ∂x ∂yIf f ( z ) is analytic , then the Cauchy - Riemann relationsare satisfied. ⇒ ∂f / ∂z * = 0 implies an analytic fonction of z containsthe combination of x + iy , not x − iy
• If Cauchy - Riemann relations are satisfied ∂ ∂u ∂ ∂v ∂ ∂v ∂ ∂u ∂ 2u ∂ 2u(1) ( ) = ( )= ( ) = − ( )⇒ 2 + 2 2 = 0 ∂x ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ∂ y ∂ 2v ∂ 2v(2) the same result for function v ( x , y ) ⇒ 2 + 2 2 =0 ∂x ∂ y⇒ u( x , y ) and v ( x , y ) are solutions of Laplace s equation in two dimension.For two families of curves u( x , y ) = conctant and v ( x , y ) = constant,the normal vectors corresponding the two curves, respectively, are  ∂u ˆ ∂u ˆ  ∂v ˆ ∂v ˆ∇u( x , y ) = i+ j and ∇v ( x , y ) = i+ j ∂x ∂y ∂x ∂y   ∂u ∂v ∂u ∂v ∂u ∂u ∂u ∂u∇u ⋅ ∇v = + =− + = 0 orthogonal ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x
• 20.3 Power series in a complex variable ∞ ∞f (z) = ∑ an z n = ∑ an r n exp(inθ ) n= 0 n=0 ∞if ∑ | an | r n is convergent ⇒ f ( z ) is absolutely convergent n=0 ∞Is ∑ | a n | r n convergent or not, can be justisfied by" Cauchy root test". n= 0 1The radius of convergence R ⇒ = lim | a n |1 / n ⇒ (1) | z |< R absolutely convergent R n→ ∞ (2) | z |> R divergent (3) | z |= R undetermined ∞ zn 1(1) ∑ ⇒ lim ( )1 / n = 0 ⇒ R = ∞ converges for all z n= 0 n! n→∞ n! ∞(2) ∑ n! z n ⇒ lim ( n! )1 / n = ∞ ⇒ R = 0 converges only at z = 0 n= 0 n →∞
• 20.4 Some elementary functions ∞ znDefine exp z = ∑ n = 0 n!Ex : Show that exp z1 exp z 2 = exp( z1 + z 2 ) ∞ ( z1 + z 2 ) nexp( z1 + z 2 ) = ∑ n=0 n! ∞ 1 n n = ∑ n!(C 0 z1 + C1n z1n−1 z2 + C 2 z1n−2 z2 + C rn z1n−r z2 + ... + C n z2 ) n 2 r n n n= 0 n s r Cr 1 n! 1set n = r + s ⇒ the coeff. of z1 z 2 is = = n! n! ( n − r )! r ! s! r ! ∞ zs ∞ zr ∞ ∞ 1 s rexp z1 exp z 2 = ∑ ∑ = ∑∑ z1 z 2 s =0 s! r = 0 r ! s =0 r = 0 s! r ! s rThere are the same coeff. of z1 z 2 for the above two terms.
• Define the complex comonent of a real number a > 0 ∞ z n (ln a ) n a = exp( z ln a ) = ∑ z n=0 n!(1) if a = e ⇒ e z = exp( z ln e ) = exp z the same as real number iy y 2 iy 3(2) if a = e, z = iy ⇒ e = exp(iy ) = 1 − − + ... 2! 3! y2 y4 y3 = 1− + + ..... + i ( y − + ...) = cos y + i sin y 2! 4! 3!(3) if a = e , z = x + iy ⇒ e x + iy = e x e iy = exp( x )(cos y + i sin y )
• Set exp w = zWrite z = r exp iθ for r is real and − π < θ ≤ π⇒ z = r exp[i (θ + 2nπ )] ⇒ w = Lnz = ln r + i (θ + 2nπ )Lnz is a multivalued function of z .Take its principal value by choosing n = 0⇒ ln z = ln r + iθ -π < θ ≤ πIf t ≠ 0 and z are both complex numbers, we define t z = exp( zLnt )Ex : Show that there are exactly n distinct nth roots of t . 1 1 tn = exp( Lnt ) and t = r exp[i (θ + 2kπ )] n 1 1 1 (θ + 2kπ ) (θ + 2kπ )⇒ tn = exp[ ln r + i ] = r n exp[i ] n n n
• 20.5 Multivalued functions and branch cutsA logarithmic function, a complex power and a complex root are allmultivalued. Is the properties of analytic function still applied?Ex : f ( z ) = z 1 / 2 and z = r exp(iθ ) (A) C y(A) z traverse any closed contour C that dose not enclose the origin, θ return r to its original value after one complete θ x circuit. (B) y(B) θ → θ + 2π enclose the origin C r 1/ 2 1/ 2 r exp(iθ / 2) → r exp[i (θ + 2π ) / 2] θ = − r 1 / 2 exp(iθ / 2) x ⇒ f (z) → − f (z) z = 0 is a branch point of the function f ( z ) = z 1 / 2
• Branch point: z remains unchanged while z traverse a closed contour Cabout some point. But a function f(z) changes after one complete circuit.Branch cut: It is a line (or curve) in the complex plane that we must cross ,so the function remains single-valued. y Ex : f ( z ) = z 1 / 2 restrict θ ⇒ 0 ≤ θ < 2π ⇒ f ( z ) is single - valued 0 x
• Ex : Find the branch points of f ( z ) = z 2 + 1 , and hence sketch suitable arrangements of branch cuts.f ( z ) = z 2 + 1 = ( z + i )( z − i ) expected branch points : z = ± iset z − i = r1 exp(iθ1 ) and z + i = r2 exp(iθ 2 )⇒ f ( z ) = r1r2 exp(iθ1 / 2) exp(iθ 2 / 2) = r1r2 exp[i (θ1 + θ 2 )]If contour C encloses(1) neither branch point, then θ1 → θ1 , θ 2 → θ 2 ⇒ f ( z ) → f ( z )(2) z = i but not z = − i , then θ1 → θ1 + 2π , θ 2 → θ 2 ⇒ f ( z ) → − f ( z )(3) z = − i but not z = i , then θ1 → θ1 , θ 2 → θ 2 + 2π ⇒ f ( z ) → − f ( z )(4) both branch points, then θ1 → θ1 + 2π ,θ 2 → θ 2 + 2π ⇒ f ( z ) → f ( z )
• f ( z ) changes value around loops containingeither z = i or z = − i . We choose branch cut as follows : (A) y (B) y i i x x −i −i
• 20.6 Singularities and zeros of complex function g( z )Isolated singularity (pole) : f ( z ) = ( z − z0 ) nn is a positive integer, g( z ) is analytic at all points insome neighborhood containing z = z0 and g( z0 ) ≠ 0,the f ( z ) has a pole of order n at z = z0 .* * An alternate definition for that f ( z ) has a pole of order n at z = z0 is lim [( z − z0 ) n f ( z )] = a z → z0 f ( z ) is analytic and a is a finite, non - zero complex number(1) if a = 0, then z = z0 is a pole of order less than n.(2) if a is infinite, then z = z0 is a pole of order greater than n.(3) if z = z0 is a pole of f ( z ) ⇒| f ( z ) |→ ∞ as z → z0(4) from any direction, if no finite n satisfies the limit ⇒ essential singularity
• Ex : Find the singularities of the function 1 1(1) f ( z ) = − 1− z 1+ z 2z⇒ f (z) = poles of order 1 at z = 1 and z = −1 (1 − z )(1 + z )(2) f ( z ) = tanh z sinh z exp z − exp(− z ) = = cosh z exp z + exp(− z ) f ( z ) has a singularity when exp z = − exp(− z )⇒ exp z = exp[i ( 2n + 1)π ] = exp(− z ) n is any integer 1⇒ 2 z = i ( 2n + 1)π ⇒ z = ( n + )πi 2Using l Hospital s rule [ z − ( n + 1 / 2)πi ] sinh z [ z − ( n + 1 / 2)πi ] cosh z + sinh z lim { }= lim { }=1z →( n+1 / 2 )πi cosh z z →( n+1 / 2 )πi sinh zeach singularity is a simple pole (n = 1)
• Remove singularties : Singularity makes the value of f ( z ) undetermined, but lim f ( z ) z→ z0 exists and independent of the direction from which z0 is approached.Ex : Show that f ( z ) = sin z / z is a removable singularity at z = 0Sol : lim f ( z ) = 0 / 0 undetermined z →0 1 z3 z5 z3 z5 f (z) = (z − + − ........) = 1 − + − .... z 3! 5! 3! 5! lim f ( z ) = 1 is independent of the way z → 0, so z →0 f ( z ) has a removable singularity at z = 0.
• The behavior of f ( z ) at infinity is given by that of f (1 / ξ ) at ξ = 0, where ξ = 1 / zEx : Find the behavior at infinity of (i) f ( z ) = a + bz −2 (ii) f ( z ) = z (1 + z 2 ) and (iii) f ( z ) = exp z(i) f ( z ) = a + bz − 2 ⇒ set z = 1 / ξ ⇒ f (1 / ξ ) = a + bξ 2 is analytic at ξ = 0 ⇒ f ( z ) is analytic at z = ∞(ii) f ( z ) = z (1 − z 2 ) ⇒ f (1 / ξ ) = 1 / ξ + 1 / ξ 3 has a pole of order 3 at z = ∞ ∞(iii) f ( z ) = exp z ⇒ f (1 / ξ ) = ∑ (n! )−1ξ −n n= 0 f ( z ) has an essential singularity at z = ∞
• If f ( z0 ) = 0 and f ( z ) = ( z − z0 )n g ( z ), if n isa positive integer, and g( z0 ) ≠ 0(i) z = z0 is called a zero of order n.(ii) if n = 1, z = z0 is called a simple zero.(iii) z = z0 is also a pole of order n of 1 / f ( z )
• 20.10 Complex integral y BA real continuous parameter t , for α ≤ t ≤ β C2x = x ( t ), y = y( t ) and point A is t = α , C1point B is t = β⇒ ∫ f ( z )dz = ∫ ( u + iv )(dx + idy ) x C C A C3 = ∫ udx − ∫ vdy + i ∫ udy + i ∫ vdx C C C C β dx β dy β dy β dx =∫ u dt − ∫ v dt + i ∫ u dt + i ∫ v dt α dt α dt α dt α dt
• Ex : Evaluate the complex integral of f ( z ) = 1 / z , along the circle |z| = R, starting and finishing at z = R. yz ( t ) = R cos t + iR sin t , 0 ≤ t ≤ 2π C1dx dy 1 x − iy R = − R sin t , = R cos t , f ( z ) = = 2 = u + iv , tdt dt x + iy x + y 2 x x cos t −y − sin tu= 2 = ,v = 2 = x + y2 R x + y2 R 1 2π cos t 2π − sin t∫C1 z dz = ∫0 R (− R sin t )dt − ∫0 ( R ) R cos tdt 2π cos t 2π − sin t + i∫ R cos tdt + i ∫ ( )( − R sin t )dt 0 R 0 R = 0 + 0 + iπ + iπ = 2πi* * The integral is also calculated by dz 2π − R sin t + iR cos t 2π ∫C1 z 0 R cos t + iR sin t =∫ dt = ∫ idt = 2πi 0The calculated result is independent of R.
• Ex : Evaluate the complex integral of f ( z ) = 1 / z along (i) the contour C 2 consisting of the semicircle | z |= R in the half - plane y ≥ 0 (ii) the contour C 3 made up of two straight lines C 3a and C 3b(i) This is just as in the previous example, but for y 0 ≤ t ≤ π ⇒ ∫ dz / z = πi iR C2 C 3b(ii) C 3a : z = (1 − t ) R + itR for 0 ≤ t ≤ 1 C 3a C 3b : − sR + i (1 − s ) R for 0 ≤ s ≤ 1 s=1 t=0 dz 1 − R + iR 1 − R − iR −R R x∫C3 z 0 R + t (− R + iR) 0 iR + s(− R − iR) =∫ dt + ∫ dt 1 −1+ i 1 2t − 1 1 11st term ⇒ ∫ dt = ∫ dt + i ∫ dt 0 1 − t + it 0 1 − 2 t + 2t 2 0 1 − 2t + 2t 2 1 i t −1/ 2 1 = [ln(1 − 2t + 2t 2 )] |1 + [2 tan −1 ( 0 )] |0 2 2 1/ 2 i π π πi a x = 0 + [ − ( − )] = ∫ a2 + x2 dx = tan −1 ( ) + c 2 2 2 2 a
• 1 1+ i 1 (1 + i )[ s − i ( s − 1)]2nd term ⇒ ∫ ds = ∫ ds 0 s + i ( s − 1) 0 2 s + ( s − 1) 2 1 2s − 1 1 1 =∫ ds + i ∫ ds 0 2s 2 − 2s + 1 0 2s 2 − 2s + 1 1 s −1/ 2 1 = [ln( 2 s 2 − 2 s + 1)] |1 + i tan −1 ( 0 ) |0 2 1/ 2 π π πi = 0 + i[ − ( − )] = 4 4 2 dz⇒ ∫C3 z = πiThe integral is independent of the different path.
• Ex : Evaluate the complex integral of f ( z ) = Re( z ) along the path C1 , C 2 and C 3 as shown in the previous examples. 2π(i) C1 : ∫ R cos t ( − R sin t + iR cos t )dt = iπR 2 0 π iπ 2(ii) C 2 : ∫ R cos t ( − R sin t + iR cos t )dt = R 0 2(iii) C 3 = C 3a + C 3b : 1 1 ∫0 (1 − t ) R( − R + iR )dt + ∫ ( − sR )( − R − iR )ds 0 1 1 = R 2 ∫ (1 − t )( −1 + i )dt + R 2 ∫ s(1 + i )ds 0 0 1 2 1 = R ( −1 + i ) + R 2 (1 + i ) = iR 2 2 2The integral depends on the different path.
• 20.11 Cauchy theoremIf f ( z ) is an analytic function, and f ( z ) is continuousat each point within and on a closed contour C⇒ ∫ f ( z )dz = 0 C ∂p( x , y ) ∂q( x , y )If and are continuous within and ∂x ∂yon a closed contour C, then by two - demensional ∂p ∂qdivergence theorem ⇒ ∫∫ ( + )dxdy = ∫ ( pdy − qdx ) R ∂x ∂y Cf ( z ) = u + iv and dz = dx + idyI = ∫ f ( z )dz = ∫ ( udx − vdy ) + i ∫ (vdx + udy ) C C C ∂ ( − u) ∂ ( − v ) ∂ ( − v ) ∂u = ∫∫ [ + ]dxdy + i ∫∫ [ + ]dxdy = 0 R ∂y ∂x R ∂y ∂xf ( z ) is analytic and the Cauchy - Riemann relations apply.
• Ex : Suppose twos points A and B in the complex plane are joined by two different paths C1 and C 2 . Show that if f ( z ) is an analytic function on each path and in the region enclosed by the two paths then the integral of f ( z ) is the same along C1 and C 2 . y B∫C 1 f ( z )dz − ∫ C2 f ( z )dz = ∫ C1 − C 2 f ( z )dz = 0 C1path C1 − C 2 forms a closed contour enclosing R R⇒∫ f ( z )dz = ∫ f ( z )dz x C1 C2 A
• Ex : Consider two closed contour C and γ in the Argand diagram, γ beingsufficiently small that it lies completely with C . Show that if the functionf ( z ) is analytic in the region between the two contours then ∫ f ( z )dz = ∫ f ( z )dz C γthe area is bounded by Γ, and f ( z ) is analytic C y∫Γ f ( z )dz = 0 γ C1 = ∫ f ( z )dz + ∫ f ( z )dz + ∫ f ( z )dz + ∫ f ( z )dz C γ C1 C2If take the direction of contour γ as that of C2contour C ⇒ ∫ f ( z )dz = ∫ f ( z )dz C γ xMorera s theorem :if f ( z ) is a continuous function of z in a closed domain Rbounded by a curve C , for ∫ f ( z )dz = 0 ⇒ f ( z ) is analytic. C
• 20.12 Cauchy’s integral formula If f ( z ) is analytic within and on a closed contour C 1 f (z) 2πi ∫C z − z0 and z0 is a point within C then f ( z0 ) = dz f (z) f (z) CI=∫ dz = ∫ dz C z − z0 γ z−z 0for z = z0 + ρ exp(iθ ), dz = iρ exp(iθ )dθ γ 2π f ( z0 + ρe iθ ) z0I=∫ iρe iθ dθ 0 ρe iθ 2π ρ →0 iθ = i∫ f ( z0 + ρe )dθ = 2πif ( z0 ) 0
• The integral form of the derivative of a complex function : 1 f (z) f ( z0 ) = 2πi ∫C ( z − z )2 dz 0 f ( z 0 + h) − f ( z 0 ) f ( z0 ) = lim h→ 0 h 1 f (z) 1 1 = lim [ h→0 2πi ∫C h z − z0 − h z − z0 )dz ] ( − 1 f (z) = lim [ h→ 0 2πi ∫C ( z − z0 − h)( z − z0 )dz ] 1 f (z) = 2πi ∫C ( z − z )2 dz 0 n! f (z) 2πi ∫C ( z − z0 )n+1For nth derivative f ( n ) ( z0 ) = dz
• Ex : Suppose that f ( z ) is analytic inside and on a circleC of radius R centered on the point z = z0 . If | f ( z ) |≤ M on the circle, where Mn! M is some constant, show that | f ( n ) ( z0 ) |≤ n . R n! f ( z )dz n! M Mn!| f ( n ) ( z0 ) |= |∫ |≤ 2πR = n 2π C ( z − z0 )n+1 2π R n+1 RLiouville s theorem : If f ( z ) is analytic and bounded for allz then f ( z ) is a constant. Mn!Using Cauchy s inequality : | f ( n ) ( z0 ) |≤ Rnset n = 1 and let R → ∞ ⇒ | f ( z0 ) |= 0 ⇒ f ( z0 ) = 0Since f ( z ) is analytic for all z , we may take z0 as anypoint in the z - plane. f ( z ) = 0 for all z ⇒ f ( z ) = constant
• 20.13 Taylor and Laurent series Taylor’s theorem: If f ( z ) is analytic inside and on a circle C of radius R centered on the point z = z0 , and z is a point inside C, then ∞ ∞ f ( n ) ( z0 ) f (z) = ∑ a n ( z − z0 ) n = ∑ n! ( z − z0 ) n n= 0 n= 0 1 f (ξ ) 2πi ∫C ξ − zf ( z ) is analytic inside and on C, so f ( z ) = dξ where ξ lies on C ∞ 1 z − z0 1 1 z − z0 nexpand ξ −z as a geometric series in ξ − z0 ⇒ = ξ − z ξ − z0 ∑( ξ − z0 ) n=0 1 f (ξ ) ∞ z − z0 n 1 ∞ f (ξ )⇒ f (z) = ∑ ( ξ − z ) dξ = 2πi ∑ ( z − z0 )n ∫C (ξ − z )n+1 dξ 2πi ∫C ξ − z0 n= 0 0 n= 0 0 1 ∞ n 2πif ( n) ( z0 ) ∞ n f (n) ( z0 ) = ∑ 2πi n=0 ( z − z0 ) n! = ∑ ( z − z0 ) n! n=0
• If f ( z ) has a pole of order p at z = z0 but is analytic at every other point insideand on C . Then g ( z ) = ( z − z0 ) p f ( z ) is analytic at z = z0 and expanded as a Taylor ∞series g( z ) = ∑ bn ( z − z0 )n . n=0Thus , for all z inside C f ( z ) can be exp anded as a Laurent series a− p a − p +1 a f (z) = + + ....... + −1 + a0 + a1 ( z − z0 ) + a 2 ( z − z0 )2 ( z − z0 ) p ( z − z0 ) p−1 z − z0 g ( n ) ( z0 ) 1 g( z )a n = bn+ p and bn = n! = 2πi ∫ ( z − z )n+1 dz C2 0 R 1 g( z ) 1 f (z) 2πi ∫ ( z − z0 ) n+1+ p 2πi ∫ ( z − z0 ) n+1⇒ an = dz = dz z 0 C1 ∞f (z) = ∑ an ( z − z0 )n is analytic in a region R between n= −∞two circles C1 and C 2 centered on z = z0
• ∞ f (z) = ∑ a n ( z − z0 ) n n = −∞(1) If f ( z ) is analytic at z = z0 , then all an = 0 for n < 0. It may happen a n = 0 for n ≥ 0, the first non - vanishing term is a m ( z-z0 ) m with m > 0, f ( z ) is said to have a zero of order m at z = z0 .(2) If f ( z ) is not analytic at z = z0 (i) possible to find a − p ≠ 0 but a − p− k = 0 for all k > 0 f ( z ) has a pole of order p at z = z0 , a −1 is called the residue of f ( z ) (ii) impossible to find a lowest value of − p ⇒ essential singularity
• 1 Ex : Find the Laurent series of f ( z ) = 3 about the singularities z ( z − 2) z = 0 and z = 2. Hence verify that z = 0 is a pole of order 1 and z = 2 is a pole of order 3, and find the residue of f ( z ) at each pole.(1) point z = 0 −1 −1 − z ( −3)( −4) − z 2 ( −3)( −4)( −5) − z 3 f (z) = 3 = [1 + ( −3)( ) + ( ) + ( ) + ...] 8 z (1 − z / 2) 8z 2 2! 2 3! 2 1 3 3 5z 2 =− − − z− − ... z = 0 is a pole of order 1 8 z 16 16 32(2) point z = 2 ⇒ set z − 2 = ξ ⇒ z( z − 2) 3 = ( 2 + ξ )ξ 3 = 2ξ 3 (1 + ξ / 2) 1 1 ξ ξ ξ ξ f (z) = 3 = 3 [1 − ( ) + ( )2 − ( )3 + ( )4 − ...] 2ξ (1 + ξ / 2) 2ξ 2 2 2 2 1 1 1 1 ξ 1 1 1 1 z−2 = − + − + − .. = − + − + − 2ξ 3 4ξ 2 8ξ 16 32 2( z − 2) 3 4( z − 2) 2 8( z − 2) 16 32z = 2 is a pole of order 3, the residue of f ( z ) at z = 2 is 1 / 8.
• How to obtain the residue ? a− m a−1f (z) = + ...... + + a0 + a1 ( z − z0 ) + a 2 ( z − z0 )2 + ... ( z − z0 ) m ( z − z0 )⇒ ( z − z0 )m f ( z ) = a − m + a − m +1 ( z − z0 ) + ....... + a −1 ( z − z0 )m −1 + ... d m −1 ∞⇒ m −1 [( z − z0 ) f ( z )] = ( m − 1)! a −1 + ∑ bn ( z − z0 )n m dz n =1Take the limit z → z0 1 d m −1R( z0 ) = a −1 = lim { [( z − z0 ) m f ( z )]} residue at z = z0 z → z0 ( m − 1)! dz m −1(1) For a simple pole m = 1 ⇒ R( z0 ) = lim [( z − z0 ) f ( z )] z → z0 g( z )(2) If f ( z ) has a simple at z = z0 and f ( z ) = , g ( z ) is analytic and h( z ) non - zero at z0 and h( z0 ) = 0 ( z − z0 ) g ( z ) ( z − z0 ) 1 g( z )⇒ R( z0 ) = lim = g ( z0 ) lim = g ( z0 ) lim = 0 z → z0 h( z ) z → z0 h( z ) z → z0 h ( z ) h ( z0 )
• Ex : Suppose that f ( z ) has a pole of order m at the point z = z0 . By considering the Laurent series of f ( z ) about z0 , deriving a general expression for the residue R( z0 ) of f ( z ) at z = z0 . Hence evaluate exp iz the residue of the function f ( z ) = 2 2 at the point z = i . ( z + 1) exp iz exp izf (z) = 2 2 = 2 2 poles of order 2 at z = i and z = − i ( z + 1) (z + i) (z − i)for pole at z = i : d d exp iz i 2 [( z − i ) 2 f ( z )] = [ 2 ]= 2 exp iz − 3 exp izdz dz ( z + i ) (z + i) (z + i) 1 i 2 −iR( i ) = [ e −1 − e −1 ] = 1! ( 2i ) 2 ( 2i ) 3 2e
• 20.14 Residue theoremf ( z ) has a pole of order m at z = z0 C ∞ γ f (z) = ∑ a n ( z − z0 ) n n= − m ρ • z0I = ∫ f ( z )dz = ∫ f ( z )dz C γset z = z0 + ρe iθ ⇒ dz = iρe iθ dθ ∞ ∞ 2πI= ∑ an ∫ ( z − z0 ) dz = C n ∑ an ∫ 0 iρ n+1e i ( n+1)θ dθ n= − m n= − m 2π n +1 i ( n+1)θ iρ n+1e i ( n+1)θ 2πfor n ≠ −1 ⇒ ∫ iρ e dθ = |0 = 0 0 i ( n + 1) 2πfor n = 1 ⇒ ∫ idθ = 2πi 0I = ∫ f ( z )dz = 2πia −1 C
• Residue theorem: f ( z ) is continuous within and on a closed contour Cand analytic, except for a finite number of poles within C ∫C f ( z )dz = 2πi ∑ R j j∑ R j is the sum of the residues of f ( z ) at its poles within C j C C
• The integral I of f ( z ) along an open contour C Cif f ( z ) has a simple pole at z = z0 ρ⇒ f ( z ) = φ ( z ) + a−1 ( z − z0 ) −1 z0φ ( z ) is analytic within some neighbour surrounding z0| z − z0 |= ρ and θ1 ≤ arg( z − z0 ) ≤ θ 2ρ is chosen small enough that no singularity of f ( z ) except z = z0I = ∫ f ( z )dz = ∫ φ ( z )dz + a −1 ∫ ( z − z0 ) −1 dz C C Climρ →0 C ∫ φ ( z )dz = 0 θ2 1I = lim ∫ f ( z )dz = lim (a −1 ∫ iθ iρe iθ dθ ) = ia −1 (θ 2 − θ1 ) ρ →0 C ρ →0 ρe θ1for a closed contour θ 2 = θ1 + 2π ⇒ I = 2πia −1
• 20.16 Integrals of sinusoidal functions 2π ∫0 F (cos θ , sin θ )dθ set z = exp iθ in unit circle 1 1 1 1 ⇒ cos θ = ( z + ), sin θ = ( z − ), dθ = − iz −1dz 2 z 2i z 2π cos 2θEx : Evaluate I = ∫ 2 2 dθ for b > a > 0 0 a + b − 2ab cos θ 1 n 1cos nθ = ( z + z − n ) ⇒ cos 2θ = ( z 2 + z − 2 ) 2 2 1 2 1 cos 2θ ( z + z − 2 )( − iz −1 )dz − ( z 4 + 1)idz dθ = 2 = 2 2a 2 + b 2 − 2ab cos θ 2 2 1 −1 z ( za 2 + zb 2 − abz 2 − ab ) a + b − 2ab ⋅ ( z + z ) 2 i ( z 4 + 1)dz i ( z 4 + 1) = = dz 2ab 2 2 a b 2ab 2 a b z ( z − z ( − + ) + 1) z ( z − )( z − ) b a b a
• i z4 + 1 2ab ∫CI= dz double poles at z = 0 and z = a / b within the unit circle a b z 2 ( z − )( z − ) b a 1 d m −1Residue : R( z0 ) = lim { [( z − z0 ) m f ( z )] z → z0 ( m − 1)! dz m −1(1) pole at z = 0, m = 2 1 d 2 z4 + 1R(0) = lim { [z 2 ]} z →0 1! dz z ( z − a / b )( z − b / a ) 4z 3 ( z 4 + 1)( −1)[2 z − (a / b + b / a )] = lim { + }= a/b+b/a z →0 ( z − a / b )( z − b / a ) 2 ( z − a / b) ( z − b / a ) 2(2) pole at z = a / b, m = 1 z4 + 1 (a / b)4 + 1 − (a 4 + b 4 )R(a / b) = lim [( z − a / b ) 2 ]= 2 = z →a / b z ( z − a / b )( z − b / a ) (a / b) (a / b − b / a ) ab(b 2 − a 2 ) i a 2 + b2 a 4 + b4 2πa 2I = 2πi × [ − ]= 2 2 2ab ab ab(b − a ) b (b − a 2 ) 2 2
• 20.17 Some infinite integrals ∞ ∫−∞ f ( x )dx f ( z ) has the following properties :(1) f ( z ) is analytic in the upper half - plane, Im z ≥ 0, except for a finite number of poles, none of which is on the real axis.(2) on a semicircle Γ of radius R, R times the maximum of y | f | on Γ tends to zero as R → ∞ (a sufficient condition is that zf ( z ) → 0 as | z |→ ∞ ). Γ 0 ∞(3) ∫ f ( x )dx and ∫0 f ( x )dx both exist −∞ x ∞ ⇒ ∫− ∞ f ( x )dx = 2π i ∑ R j −R 0 R jfor | ∫ f ( z )dz |≤ 2π R × (maximum of | f | on Γ ), the integral along Γ Γtends to zero as R → ∞ .
• ∞ dx Ex : Evaluate I = ∫0 2 (x + a ) 2 4 a is real dz R dx dz∫C ( z 2 + a 2 )4 = ∫− R ( x 2 + a 2 )4 ∫Γ ( z 2 + a 2 )4 + as R → ∞ Γ dz dz ∞ dx ai⇒∫ →0⇒ ∫C ( z 2 + a 2 )4 = ∫− ∞ ( x 2 + a 2 )4 Γ ( z 2 + a 2 )4( z 2 + a 2 )4 = 0 ⇒ poles of order 4 at z = ± ai , −R 0 Ronly z = ai at the upper half - plane 1 1 1 iξ − 4set z = ai + ξ , ξ → 0 ⇒ 2 2 4 = 2 4 = 4 (1 − ) (z + a ) ( 2aiξ + ξ ) ( 2aiξ ) 2a 1 ( −4)( −5)( −6) − i 3 − 5ithe coefficient of ξ −1 is ( ) = ( 2a ) 4 3! 2a 32a 7 ∞ dx − 5i 10π 1 10π 5π∫0 ( x 2 + a 2 )4 = 2πi ( 32a 7 )= 32a 7 ⇒I = × 2 32a 7 = 32a 7
• For poles on the real axis: y ΓPrincipal value of the integral, defined as ρ → 0 R z0 − ρ RP∫ f ( x )dx = ∫− R f ( x )dx + ∫z + ρ f ( x )dx γ ρ −R 0 -R 0 z0 R xfor a closed contour C z0 − ρ R∫C f ( z )dz = ∫− R f ( x )dx + ∫γ f ( z )dz + ∫z + ρ 0 f ( x )dx + ∫Γ f ( z )dz R = P∫ f ( x )dx + ∫γ f ( z )dz + ∫Γ f ( z )dz −R(1) for ∫ f ( z )dz has a pole at z = z0 ⇒ ∫ f ( z )dz = −πia1 γ γ iθ(2) for ∫ f ( z )dz set z = Re dz = i Re iθ dθ Γ ⇒ ∫Γ f ( z )dz = ∫Γ f (Re iθ )i Re iθ dθIf f ( z ) vanishes faster than 1 / R 2 as R → ∞, the integral is zero
• Jordan’s lemma(1) f ( z ) is analytic in the upper half - plane except for a finite number of poles in Im z > 0( 2) the maximum of | f ( z ) |→ 0 as | z |→ ∞ in the upper half - plane(3) m > 0, then ∫Γ e imz IΓ = f ( z )dz → 0 as R → ∞, Γ is the semicircular contourfor 0 ≤ θ ≤ π / 2, 1 ≥ sin θ / θ ≥ π / 2 f (θ ) f = 2θ / π| exp(imz )| = | exp(− mR sin θ )| πIΓ ≤ ∫Γ |e imz f ( z )||dz| ≤ MR ∫ e − mR sin θ dθ 0 f = sin θ π/ 2 − mR sin θ = 2 MR ∫ e dθ 0M is the maximum of |f ( z )| on |z| = R, R → ∞ M → 0 π /2 θ πM π / 2 − mR ( 2θ / π ) πMI Γ < 2 MR ∫ e (1 − e − mR ) < dθ = 0 m mas R → ∞ ⇒ M → 0 ⇒ I Γ → 0
• cos mx ∞Ex : Find the principal value of ∫− ∞ x − a dx a real, m > 0 Γ e imzConsider the integral I = ∫C z−a dz = 0 no pole in the γ ρ −1upper half - plane, and |( z − a ) | → 0 as |z| → ∞ -R 0 a R e imzI = ∫C z−a dz a−ρ e imx e imz R e imx e imz = ∫− R x−a dx + ∫γ z−a dz + ∫a + ρ x−a dx + ∫Γ z−a dz = 0 e imzAs R → ∞ and ρ → 0 ⇒ ∫ dz → 0 Γ z−a ∞e imx⇒ P∫ dx − iπa−1 = 0 and a−1 = e ima −∞ x − a ∞ cos mx ∞ sin mx⇒ P∫ dx = −π sin ma and P ∫ dx = π cos ma −∞ x − a −∞ x − a
• 20.18 Integral of multivalued functions 1/ 2 yMultivalued functions such as z , LnzSingle branch point is at the otigin. We let R → ∞ R Γand ρ → 0. The integrand is multivalued, its values γalong two lines AB and CD joining z = ρ to z = R A Bare not equal and opposite. ρ x C D ∞ dx Ex : I = ∫0 ( x + a )3 x1 / 2 for a > 0(1) the integrand f ( z ) = ( z + a ) −3 z −1 / 2 , |zf ( z )| → 0 as ρ → 0 and R → ∞ the two circles make no contribution to the contour integral(2) pole at z = − a , and ( − a )1 / 2 = a 1 / 2 e iπ / 2 = ia 1 / 2 1 d 3 −1 1 R( − a ) = lim [( z + a )3 ] z → − a ( 3 − 1)! dz 3 − 1 3 1/ 2 (z + a) z 1 d 2 −1 / 2 − 3i = lim z = z → − a 2! dz 2 8a 5 / 2
• − 3i∫AB dz + ∫Γ dz + ∫DC dz + ∫γ dz = 2πi ( 8a 5/2 )and ∫ dz = 0 and ∫γ dz = 0 Γalong line AB ⇒ z = xe i 0 , along line CD ⇒ z = xe i 2π ∞ dx 0 dx 3π∫0, A→ B ( x + a )3 x1 / 2 ∞,C → D ( xe i 2π + a )3 x1 / 2e (1 / 2×2πi ) 4a 5 / 2 +∫ = 1 ∞ dx 3π⇒ (1 − iπ )∫ = e 0 ( x + a )3 x1 / 2 4a 5 / 2 ∞ dx 3π⇒∫ = 0 ( x + a )3 x1 / 2 8a 5 / 2
• ∞ x sin x Ex : Evaluate I (σ ) = ∫− ∞ x 2 − σ 2 dx z sin z 1 ze iz 1 ze − iz ∫C z2 − σ 2 dz = 2i ∫C 1 z2 − σ 2 dz − 2i ∫C 2 2 z −σ 2 dz = I1 + I 2(1) for I1 , the contour is choosed on the upper half - plane C1 due to the term e iz , and only one pole at z = σ . Γ 1 ze iz 1 −σ − ρ xe ix γ1I1 = 2i ∫C 1 z2 − σ 2 dz = 2i ∫− R x2 − σ 2 dx ρ 1 σ −ρ xe ix 1 ∞ xe ix -R -σ σ R + 2i ∫−σ + ρ x2 − σ 2 dx + 2i ∫σ + ρ x 2 − σ 2 dx γ2 1 ze iz 1 ze iz 1 ze iz + 2i ∫γ 1 z2 − σ 2 dz + ∫ 2 2i γ 2 z − σ 2 dz + 2i ∫Γ z 2 − σ 2 dz 1 σe iσ π = 2πi × Res( z = σ ) = π = e iσ 2i 2σ 2
• As ρ → 0 and R → ∞ ⇒ ∫ dz → 0 Γ1 ze iz 1 − π − iσ2i ∫γ 1 ( z + σ )( z − σ ) dz = 2i × ( −πi )Res( z = −σ ) = 4 e1 ze iz 1 π ∫γ dz = × πiRes( z = σ ) = e iσ2i 2 ( z + σ )( z − σ ) 2i 4 1 ∞ xe ix π π γ1 dx + (e iσ − e − iσ ) = e iσ 2i ∫− ∞ x 2 − σ 2I1 = σ 4 2 −σ(2) for I 2 , we choose the lower half - plane by the − iz Γ γ2 term e , only one pole at z = −σ −1 ze − iz − 1 −σ − ρ xe − ixI2 = 2i ∫C2 z 2 − σ 2 dz = 2i ∫− R x 2 − σ 2 dx C2 1 σ −ρ xe − ix 1 ∞ xe − ix 1 ze − iz − 2i ∫−σ + ρ x2 − σ 2 dx − 2i ∫σ + ρ x2 − σ 2 dx − 2i ∫γ 1 2 z −σ 2 dz 1 ze − iz 1 ze − iz −1 ( −σ )e iσ π − ∫γ dz − ∫Γ dz = ( ) × ( −2πi ) = e iσ 2i 2 z2 − σ 2 2i z2 − σ 2 2i − 2σ 2
• As ρ → 0, R → ∞ ⇒ ∫ dz → 0 Γ−1 ze − iz −1 ( −σ )e iσ π = e iσ2i ∫γ 1 ( z + σ )( z − σ ) dz = ( )( −πi ) 2i − 2σ 4−1 ze − iz −1 σe − iσ − π − iσ2i ∫γ 2 ( z + σ )( z − σ )dz = ( 2i )(πi ) 2σ = 4 e − 1 ∞ xe − ix π π dx + (e iσ − e − iσ ) = e iσ 2i ∫− ∞ x 2 − σ 2I2 = 4 2 − 1 ∞ xe − ix π 1 dx = e iσ − (e iσ − e − iσ ) 2i ∫− ∞ x 2 − σ 2⇒ 2 4 ∞ x sin x 1 ∞ xe ix 1 ∞ xe − ixI (σ ) = ∫ 2i ∫− ∞ x 2 − σ 2 2i ∫− ∞ x 2 − σ 2 dx = dx − dx −∞ x 2 − σ 2 π π π π = e iσ − ( e iσ − e − iσ ) + e iσ − ( e iσ − e − iσ ) 2 4 2 4 π π π = πe iσ − e iσ + e − iσ = (e iσ + e − iσ ) = π cos σ 2 2 2