Cauchy's Integral Theorem | Mathquery

Cauchy's Integral Theorem:

Statement :

      The theorem is usually formulated for closed paths as follows :

       Let U be an open subset of C which is simply connected . Let f: U-->C be a holomorphic  function , and let γ be a rectifiable path in U whose start point is equal to its end point . Then

                        ∮ f(z) dz = 0
                        γ

Proof :

      Let us assume that the  partial derivatives  of a holomorphic function are continuous , the Cauchy Integral Theorem can be proved as direct sequence of    Green's Theorem and the fact that the real and imaginary parts of f = u + i v must satisfy the Cauchy - Riemann equations in the region bounded by γ , and moreover in the open neighbourhood U of this region.

             Cauchy provided this proof , but it was later proved by Goursat without requiring techniques from vector calculus , or the continuity of partial derivatives .

            We can break the integrand f , as well as the differential dz into their real and imaginary components .

                    f = u + i v   ..............(1)

                  dz = dx + i dy .............(2)

            Then we have 

 ∮ f(z) dz = ∮ (u+iv)(dx+idy)
 γ                γ

                 = ∮(u dx - v dy)+i∮(v dx + u dy)
                    γ                         γ             ........(3)

      By Green's Theorem , we then replace the integrals around the closed contour γ with an area integral throughout the domain D that is enclosed by γ as follows.

 ∮(u dx - v dy) = ∫∫(-∂v/∂x - ∂u/∂y) dx dy
 γ                            D                          ............(4)

 ∮(v dx + u dy) = ∫∫(∂u/∂x - ∂v/∂y) dx dy
 γ                             D                         ............(5)

  But the real and imaginary parts of a function holomorphic in the domain D , u and v must satisfy the Cauchy-Riemann equations so ,

               ∂u/∂x = ∂v/∂y ..............(6)
               ∂u/∂y = -∂v/∂x .............(7)

    We therefore find that both integrands are zero

     ∫∫(-∂v/∂x - ∂u/∂y) dx dy
      D

                             = ∫∫(∂u/∂y - ∂u/∂y) dx dy =0
                                  D                      ...........(8)

     ∫∫(∂u/∂x - ∂v/∂y) dx dy
      D
                            = ∫∫(∂u/∂x - ∂u/∂x) dx dy
                                 D                      .............(9)

 This gives the desired result

                     ∮ f(z) dz = 0           ...............(10)
                    γ                                (Proved)

Example :

       Prove that ∫ eᶻ dz = 0

                          c

Solution :

      Here f(z) = eᶻ = e^(x+iy)   (∵ In complex 

                                                       z = x+iy)

                               = e^x . e^ iy

                               = e^x (cos y + i sin y)

                                              (∵ polar form of 

                                               complex function)

                              = e^x cos y + i e^x sin y

                              = u + iv

where   u = e^x cosy           v= e^x sin y

      ∂u/∂x = e^x cosy           ∂v/∂x = e^x sin y

      ∂u/∂y = -e^x sin y          ∂v/∂y = e^x cos y

∴   ∂u/∂x = ∂v/∂y    and      ∂u/∂y = - ∂v/∂x

  ⇒ f(z) = eᶻ  satisfy C-R equation

  ⇒  eᶻ is analytic for all z .

 So ∫ eᶻ dz = 0 by Cauchy's Integral Theorem

      c

  Which is the required solution .

About Scientist :-

         Baron Augustin-Louis Cauchy (21 August 1789 – 23 May 1857) was a French mathematician, engineer, and physicist who made pioneering contributions to several branches of mathematics, including mathematical analysis and continuum mechanics .For More