f (z) dz = 0 f(z) dz = 2πj f(z 0 ) Generalized Cauchy Integral Formula (For pole with any order) (n 1)! f (n 1) (z 0 ) f (n) (z 0 ) M n!

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1 uhy s Theorems I Ang M.S. Otober 26, 212 Augustin-Louis uhy Referenes Murry R. Spiegel omplex V ribles with introdution to onf orml mpping nd its pplitions Dennis G. Zill, P. D. Shnhn A F irst ourse in omplex Anlysis with Applitions J. H. Mthews, R. W. Howell omplex Anlysis F or Mthemtis nd Engineerng 1 Summry uhy Integrl Theorem Let f be nlyti in simply onneted domin D. If is simple losed ontour tht lies in D, nd there is no singulr point inside the ontour, then = uhy Integrl Formul (For simple pole) If there is singulr point z inside the ontour, then f(z) z z = j f(z ) Generlized uhy Integrl Formul (For pole with ny order) uhy Inequlity Guss Men Vlue Theorem f(z) (z z ) n = j (n 1)! f (n 1) (z ) f(z ) = 1 f (n) (z ) M n! r n f ( z + re jθ) dθ 1

2 2 Relted Mthemtis Review 2.1 Stoke s Theorem (The proof is skipped) F ds = F dr onsider F = (F X, F Y, F Z ) Let F Z = x ŷ ẑ F = det x z F X F Y F Z x ŷ ẑ F = det x z F X F Y ds = nds, for ds = dxdy, n = ẑ. By ẑ ẑ = 1, onsider ẑ omponent only for F x ŷ ẑ ( F = det x z = FY x F ) X ẑ F X F Y i.e. onsider the RHS (let F Z = ) F dr = Thus F ds = 2.2 uhy-reimnn ondition ( FY x F ) X dxdy (F X, F Y, F Z ) (dx, dy, ) = ( FY x F ) X dxdy = F X dx + F Y dy F X dx + F Y dy For f(z) = u(z) + jv(z) = u(x, y) + jv(x, y) The omplex derivtive is If the limit exists. lim h h f (z + h) f(z ) h = f (z ) 2

3 If the limit exists, limits long rel xis should be the sme the limit long imginry xis lim h h R f (z + h) f (z ) h = f (z ) x = 1 j f (z ) = lim h h R f (z + jh) f (z ) jh i.e. The uhy-riemnn Eqution Expnd f (z ) x = 1 j f (z ) Equlize rel prt nd imginry prt Or simply s x [u (x, y ) + jv (x, y )] = 1 j [u (x, y ) + jv (x, y )] x u (x, y ) = v (x, y ) x v (x, y ) = u (x, y ) u x = v v x = u i.e. If f(z) is nlyti, then it fulfill this ondition 2.3 ML Inequlity First, This is true if if g(x) f(x), then b g(x)dx b f(x)dx sup g(x) sup f(x) Then for ny funtion f, it is true tht inf g(x) inf f(x) Then pply the inequlity bove, f f f b f(x) dx b f(x)dx b Now is the time to show, for bounded f(z), i.e. f(z) M, f(z) ML 3 f(x) dx

4 Pf. By using the b f(x)dx b f(x) dx f(z) f(z) Sine f(z) si bounded, f(z) f(z) M = M = ML }{{} Therefore f(z) ML L 3 The uhy Theorem f(z) = u(x, y) + jv(x, y) = [u (x, y) + jv (x, y)] [dx + jdv] = udx vdy + j vdx + udy 3.1 The Rel Prt Let F X = u (x, y) F Y = u (x, y) nd onsider the rel prt of the integrl udx vdy = F X dx + F Y dy By the Stoke Theorem udx + ( v)dy = [( v) x u y ] dxdy = [ v x u y ] dxdy Sine the funtion is nlyti in the region D, the funtion fulfill uhy-reimnn ondition v x = u Then [ v x u y ] dxdy = [u y u y ] dxdy = The rel prt of f(z) equl zero 4

5 3.2 The imginry prt onsider the imginry prt of Let F X be v(x, y) nd F Y f(z), vdx + udy be u(x, y), nd pply Stoke s Theorem vdx + udy = (u x v y ) dxdy As the funtion is nlyti, so it fulfill uhy-reimnn ondition : beome zero u x = v, thus the integrl Finlly, Let f be nlyti in simply onneted domin D, for the simple losed ontour tht lies in D, the lose ontour integrl equl zero = 4 The uhy Integrl Formuls 4.1 Simple Pole ondition g (z) z z = jg (z ) z inside, if z is outside, then the integrl is just zero is nlyti in simply onneted domin domin of is simply onnet losed region, nd fulfill R-ondition there Proof. onsider the digrm By onept of pth deformtion, the integrl of z z long pth is equivlent to the integrl long pth ϵ tht ϵ = lim z z ϵ ϵ z z The smll irle inside n be expressed s As the rdius ϵ +, the irle is thus z z = ϵe jθ θ < z z = ϵe jθ θ < 5

6 Put this bk into the integrl = j = lim z z ϵ ϵ = lim ϵ = lim z z ϵ g ( z + ϵe jθ) ϵe jθ [ lim ϵ g ( z + ϵe jθ)] dθ = j g ( z + ϵe jθ) (z + ϵe jθ ) z d ( z + ϵe jθ) jϵe jθ dθ = jlim g ( z + ϵe jθ) dθ ϵ g (z ) dθ = jg (z ) dθ = jg (z ) z z = jg (z ) 4.2 Generlized uhy Integrl Formul f (z ) (z z ) n = j (n 1)! j f (n 1) (z ) First, the uhy Integrl Formul for simple pole is g (z) = jg (z ) z z Tke d on both side (Not d )! d g (z) = j d g (z) j g (z ) 2 = z z (z z ) 1! g (z ) Repet, d d g (z) j d 2 = g (z ) (z z ) 1! g (z) j d 3 = g (z ) (z z ) 1! Thus, the generl form of uhy Integrl Formul is g (z) (z z ) n = j (n 1)! g(n 1) (z ) This n be proved using Mthemtil Indution. g (z) j 3 = (z z ) 2! g (z ) g (z) j 4 = (z z ) 3! g(3) (z ) 6

7 5 onsequenes of uhy s Integrl Formul 5.1 uhy Inequlity The generlized uhy Integrl Formul is j = (z z ) n! f (n) (z ) Tke bsolute vlue (z z ) Rerrnge Sine = j n! f (n) (z ) f (n) (z ) = n! (z z ) (z z ) n! = j n! (z z ) f (n) (z ) z = z + re jθ (z z ) = re jθ (z z ) = r e j()θ = r }{{} 1 And pply ML inequlity f (n) (z ) n! r n! r Finlly 5.2 Guss Men Vlue Theorem f (n) (z ) M n! r n M = M n! r }{{} r = M n! r r Rell, the uhy Integrl Formul for simple pole = j f (z ) z z Rerrnge, nd tke z = z + re jθ, = rje jθ dθ f (z ) = 1 j nel out ommon ftor = 1 f ( z + re jθ) ( rje jθ dθ ) z z j re jθ f (z ) = 1 f ( z + re jθ) dθ END 7