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Cauchy Integral Theorem Cauchy’s Theorems I Ang M.S. Augustin-Louis Cauchy October 26, 2012 1789 – 1857 References Murray R. Spiegel Complex V ariables with introduction to conformal mapping and its applications Dennis G. Zill , P. D. Shanahan A F irst Course in Complex Analysis with Applications J. H. Mathews , R. W. Howell Complex Analysis F or Mathematics and Engineerng 1 Summary • Cauchy Integral Theorem Let f be analytic in a simply connected domain D. If C is a simple closed contour that lies in D , and there is no singular point inside the contour, then ˆ f (z) dz = 0 C • Cauchy Integral Formula (For simple pole) If there is a singular point z0 inside the contour, then ˛ f(z) dz = 2πj f(z0) z − z0 • Generalized Cauchy Integral Formula (For pole with any order) ˛ f(z) 2πj (n−1) n dz = f (z0) (z − z0) (n − 1)! • Cauchy Inequality M · n! f (n)(z ) ≤ 0 rn • Gauss Mean Value Theorem ˆ 2π ( ) 1 jθ f(z0) = f z0 + re dθ 2π 0 1 2 Related Mathematics Review 2.1 Stoke’s Theorem ¨ ˛ r × F¯ · dS = F¯ · dr Σ @Σ (The proof is skipped) ¯ Consider F = (FX ;FY ;FZ ) 0 1 x^ y^ z^ B @ @ @ C r × F¯ = det B C @ @x @y @z A FX FY FZ Let FZ = 0 0 1 x^ y^ z^ B @ @ @ C r × F¯ = det B C @ @x @y @z A FX FY 0 dS =ndS ^ , for dS = dxdy , n^ =z ^ . By z^ · z^ = 1 , consider z^ component only for r × F¯ 0 1 x^ y^ z^ ( ) B @ @ @ C @F @F r × F¯ = det B C = Y − X z^ @ @x @y @z A @x @y FX FY 0 i.e. ¨ ¨ ( ) @F @F r × F¯ · dS = Y − X dxdy Σ Σ @x @y Consider the RHS (let FZ = 0) ˛ ˛ ˛ ¯ F · dr = (FX ;FY ;FZ ) · (dx; dy; dz) = FX dx + FY dy @Σ @Σ @Σ Thus ¨ ( ) ˛ @FY @FX − dxdy = FX dx + FY dy Σ @x @y @Σ 2.2 Cauchy-Reimann Condition For f(z) = u(z) + jv(z) = u(x; y) + jv(x; y) The complex derivative is f (z0 + h) − f(z0) 0 lim = f (z0) h ! 0 h h 2 C If the limit exists. 2 If the limit exists, limits along real axis should be the same the limit along imaginary axis f (z + h) − f (z ) @f (z ) 1 @f (z ) f (z + jh) − f (z ) lim 0 0 = 0 = 0 = lim 0 0 h ! 0 h @x j @y h ! 0 jh h 2 R h 2 R i.e. The Cauchy-Riemann Equation @f (z ) 1 @f (z ) 0 = 0 @x j @y Expand @ 1 @ [u (x ; y ) + jv (x ; y )] = [u (x ; y ) + jv (x ; y )] @x 0 0 0 0 j @y 0 0 0 0 Equalize real part and imaginary part 8 > @ @ < u (x ; y ) = v (x ; y ) @x 0 0 @y 0 0 > @ @ : v (x ; y ) = − u (x ; y ) @x 0 0 @y 0 0 Or simply as 8 > @u @v < = @x @y > @v @u : = − @x @y i.e. If f(z) is analytic, then it fulfill this condition 2.3 ML Inequality First, ˆ ˆ b b if g(x) ≤ f(x) , then g(x)dx ≤ f(x)dx a a This is true if sup g(x) ≤ sup f(x) inf g(x) ≤ inf f(x) Then for any function f , it is true that −|fj ≤ f ≤ jfj Then apply the inequality above , ˆ ˆ ˆ b b b − jf(x)jdx ≤ f(x)dx ≤ jf(x)jdx a a a Now is the time to show , for bounded f(z) , i.e. jf(z)j ≤ M , ˆ f(z)dz ≤ ML c 3 Pf. ´ ´ b ≤ b j j By using the a f(x)dx a f(x) dx ˆ ˆ f(z)dz ≤ jf(z)jdz c c Since f(z) si bounded, ˆ ˆ ˆ ˆ f(z)dz ≤ jf(z)jdz ≤ Mdz = M dz = ML c c c | C{z } L Therefore ˆ j f(z)dzj ≤ ML c 3 The Cauchy Theorem f(z) = u(x; y) + jv(x; y) ˛ ˛ ˛ ˛ f (z) dz = [u (x; y) + jv (x; y)] [dx + jdv] = udx − vdy + j vdx + udy C C C C 3.1 The Real Part Let FX = u (x; y) FY = −u (x; y) and consider the real part of the integral ˛ ˛ udx − vdy = FX dx + FY dy C By the Stoke’ Theorem ˛ ¨ ¨ − − − − − udx + ( v)dy = [( v)x uy] dxdy = [ vx uy] dxdy C Since the function is analytic in the region D , () the function fulfill Cauchy-Reimann Condition @v @u = − @x @y Then ¨ ¨ [−vx − uy] dxdy = [uy − uy] dxdy = 0 ´ The real part of C f(z)dz equal zero 4 3.2 The imaginary part ´ Consider the imaginary part of C f(z)dz , ˛ vdx + udy C Let FX be v(x; y) and FY be u(x; y) , and apply Stoke’s Theorem ˛ ¨ vdx + udy = (ux − vy) dxdy C @u @v As the function is analytic, so it fulfill Cauchy-Reimann Condition : = , thus the integral @x @y become zero Finally, Let f be analytic in a simply connected domain D , for the simple closed contour C that lies in D , the close contour integral equal zero ˆ f (z) dz = 0 C 4 The Cauchy Integral Formulas 4.1 Simple Pole ˛ g (z) − dz = 2πjg (z0) C z z0 Condition • z0 inside C , if z0is outside C , then the integral is just zero • g(z) is analytic in simply connected domain () domain of g(z) is simply connect closed region, and g(z) fulfill CR-Condition there Proof. Consider the diagram g(z) By concept of path deformation , the integral of z − z0 along path C is equivalent to the integral along path Cϵ that ϵ ! 0 ˛ ˛ g(z) g(z) dz = lim dz − ϵ!0 − C z z0 Cϵ z z0 The small circle inside can be expressed as jθ jz − z0j = ϵe 0 ≤ θ < 2π As the radius ϵ ! 0+ , the circle is thus jθ z − z0 = ϵe 0 ≤ θ < 2π 5 Put this back into the integral ˛ ˛ ˆ ( ) 2π jθ ( ) g(z) g(z) g z0 + ϵe jθ dz = lim dz = lim d z0 + ϵe − ϵ!0 − ϵ!0 jθ − C z z0 Cϵ z z0 0 (z0 + ϵe ) z0 ˆ ( ) ˆ 2π g z + ϵejθ 2π ( ) = lim 0 jϵejθdθ = jlim g z + ϵejθ dθ ! jθ ! 0 ϵ 0 0 ϵe ϵ 0 0 ˆ ˆ ˆ 2π h ( )i 2π 2π = j limg z + ϵejθ dθ = j g (z ) dθ = jg (z ) dθ = 2πjg (z ) ! 0 0 0 0 0 ϵ 0 0 0 ) ˛ g(z) − dz = 2πjg (z0) C z z0 4.2 Generalized Cauchy Integral Formula ˛ f (z0) 2πj (n−1) n dz = 2πj f (z0) (z − z0) (n − 1)! First, the Cauchy Integral Formula for simple pole is ˛ g (z) dz = 2πjg (z ) z − z 0 ( ) C 0 d d Take on both side Not ! dz0 ˛ dz ˛ d g (z) d g (z) 2πj dz = 2πj g (z ) ) dz = g0 (z ) − 0 2 0 dz0 C z z0 dz0 C (z − z0) 1! Repeat, ˛ ˛ d g (z) 2πj d 0 ) g (z) 2πj 2 dz = g (z0) 3 dz = g"(z0) dz0 C (z − z0) 1! dz0 C (z − z0) 2! ˛ ˛ d g (z) 2πj d g (z) 2πj ) (3) 3 dz = g"(z0) 4 dz = g (z0) dz0 C (z − z0) 1! dz0 C (z − z0) 3! Thus, the general form of Cauchy Integral Formula is ˛ g (z) 2πj (n−1) − n dz = − g (z0) C (z z0) (n 1)! This can be proved using Mathematical Induction. 6 5 Consequences of Cauchy’s Integral Formula 5.1 Cauchy Inequality The generalized Cauchy Integral Formula is ˛ f (z) 2πj (n) n+1 dz = f (z0) C (z − z0) n! Take absolute value ˛ ˛ f (z) 2πj f (z) j2πj · jjj dz = f (n) (z ) ) dz = f (n) (z ) n+1 0 n+1 j j 0 C (z − z0) n! C (z − z0) n! Rearrange ˛ ˛ n! f (z) n! f (z) (n) ≤ f (z0) = n+1 dz n+1 dz 2π C (z − z0) 2π C (z − z0) Since jθ ) − jθ ) − n+1 j n+1j · j j(n+1)θj n+1 z = z0 + re (z z0) = re (z z0) = r |e {z } = r 1 And apply ML inequality ˛ ˛ ˛ n! f (z) n! M · n! M · n! f (n) (z ) ≤ dz ≤ Mdz = dz = · 2πr 0 n+1 · n+1 · n+1 · n+1 2π C r 2π r C 2π r | C{z } 2π r 2πr Finally M · n! f (n) (z ) ≤ 0 rn 5.2 Gauss Mean Value Theorem Recall, the Cauchy Integral Formula for simple pole ˛ f (z) − dz = 2πj f (z0) C z z0 jθ jθ Rearrange, and take z = z0 + re , dz = rje dθ ˛ ˆ ( ) 1 f (z) 1 2π f z + rejθ ( ) f (z ) = dz = 0 rjejθdθ 0 − jθ 2πj C z z0 2πj 0 re Cancel out common factor ˆ 2π ( ) 1 jθ f (z0) = f z0 + re dθ 2π 0 END 7.
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