Let $f (z) = u+iv$ be an analytic function, then show that $ (∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 =...












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Let $f (z) = u+iv$ be an analytic function, then show that $$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2,.$$




$f(z) = u + iv $



$ϕ = |f(z)|^2 = u^2 + v^2 $



$f'(z) = ∂u/∂x + i∂v/∂x $



$|f'(z)|^2 = (∂u/∂x)^2 + (∂v/∂x)^2 = (∂u/∂y)^2 + (∂v/∂y)^2 $



$∂ϕ/∂x = 2u∂u/∂x + 2v∂v/∂x $



$∂^2ϕ/∂x^2 = (∂u/∂x)^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + (∂v/∂x)^2 $



$∂^2ϕ/∂x^2 = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) $ ----(1)



Similarly for,



$∂^2ϕ/∂y^2 = 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $ ----(2)



(1)+(2),



$(∂^2ϕ/∂x^2 + ∂^2ϕ/∂y^2) = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



How it that the extra terms remain ??






complex-analysis derivatives partial-derivative absolute-value holomorphic-functions







share|cite|improve this question











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    $begingroup$



    Let $f (z) = u+iv$ be an analytic function, then show that $$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2,.$$




    $f(z) = u + iv $



    $ϕ = |f(z)|^2 = u^2 + v^2 $



    $f'(z) = ∂u/∂x + i∂v/∂x $



    $|f'(z)|^2 = (∂u/∂x)^2 + (∂v/∂x)^2 = (∂u/∂y)^2 + (∂v/∂y)^2 $



    $∂ϕ/∂x = 2u∂u/∂x + 2v∂v/∂x $



    $∂^2ϕ/∂x^2 = (∂u/∂x)^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + (∂v/∂x)^2 $



    $∂^2ϕ/∂x^2 = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) $ ----(1)



    Similarly for,



    $∂^2ϕ/∂y^2 = 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $ ----(2)



    (1)+(2),



    $(∂^2ϕ/∂x^2 + ∂^2ϕ/∂y^2) = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



    $(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



    How it that the extra terms remain ??






    complex-analysis derivatives partial-derivative absolute-value holomorphic-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $f (z) = u+iv$ be an analytic function, then show that $$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2,.$$




      $f(z) = u + iv $



      $ϕ = |f(z)|^2 = u^2 + v^2 $



      $f'(z) = ∂u/∂x + i∂v/∂x $



      $|f'(z)|^2 = (∂u/∂x)^2 + (∂v/∂x)^2 = (∂u/∂y)^2 + (∂v/∂y)^2 $



      $∂ϕ/∂x = 2u∂u/∂x + 2v∂v/∂x $



      $∂^2ϕ/∂x^2 = (∂u/∂x)^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + (∂v/∂x)^2 $



      $∂^2ϕ/∂x^2 = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) $ ----(1)



      Similarly for,



      $∂^2ϕ/∂y^2 = 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $ ----(2)



      (1)+(2),



      $(∂^2ϕ/∂x^2 + ∂^2ϕ/∂y^2) = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



      $(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



      How it that the extra terms remain ??






      complex-analysis derivatives partial-derivative absolute-value holomorphic-functions







      share|cite|improve this question











      $endgroup$





      Let $f (z) = u+iv$ be an analytic function, then show that $$(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2,.$$




      $f(z) = u + iv $



      $ϕ = |f(z)|^2 = u^2 + v^2 $



      $f'(z) = ∂u/∂x + i∂v/∂x $



      $|f'(z)|^2 = (∂u/∂x)^2 + (∂v/∂x)^2 = (∂u/∂y)^2 + (∂v/∂y)^2 $



      $∂ϕ/∂x = 2u∂u/∂x + 2v∂v/∂x $



      $∂^2ϕ/∂x^2 = (∂u/∂x)^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + (∂v/∂x)^2 $



      $∂^2ϕ/∂x^2 = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) $ ----(1)



      Similarly for,



      $∂^2ϕ/∂y^2 = 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $ ----(2)



      (1)+(2),



      $(∂^2ϕ/∂x^2 + ∂^2ϕ/∂y^2) = 2|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2|f'(z)|^2 + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



      $(∂^2/∂x^2 + ∂^2/∂y^2)|f(z)|^2 = 4|f'(z)|^2 + 2u(∂^2u/∂x^2) + 2v(∂^2v/∂x^2) + 2u(∂^2u/∂y^2) + 2v(∂^2v/∂y^2) $



      How it that the extra terms remain ??






      complex-analysis derivatives partial-derivative absolute-value holomorphic-functions




      complex-analysis derivatives partial-derivative absolute-value holomorphic-functions






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 13 '18 at 8:57









      Batominovski

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      33.2k33293










      asked Dec 13 '18 at 8:19









      The IdiotThe Idiot

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          Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$frac {partial ^{2}} {partial x^{2}} u+ frac {partial ^{2}} {partial y^{2}}u=0$ and $frac {partial ^{2}} {partial x^{2}} v+ frac {partial ^{2}} {partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.






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            Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$frac {partial ^{2}} {partial x^{2}} u+ frac {partial ^{2}} {partial y^{2}}u=0$ and $frac {partial ^{2}} {partial x^{2}} v+ frac {partial ^{2}} {partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.






            share|cite|improve this answer











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              $begingroup$

              Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$frac {partial ^{2}} {partial x^{2}} u+ frac {partial ^{2}} {partial y^{2}}u=0$ and $frac {partial ^{2}} {partial x^{2}} v+ frac {partial ^{2}} {partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.






              share|cite|improve this answer











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                $begingroup$

                Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$frac {partial ^{2}} {partial x^{2}} u+ frac {partial ^{2}} {partial y^{2}}u=0$ and $frac {partial ^{2}} {partial x^{2}} v+ frac {partial ^{2}} {partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.






                share|cite|improve this answer











                $endgroup$



                Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$frac {partial ^{2}} {partial x^{2}} u+ frac {partial ^{2}} {partial y^{2}}u=0$ and $frac {partial ^{2}} {partial x^{2}} v+ frac {partial ^{2}} {partial y^{2}}v=0$. You can prove this easily using C-R equations]. Hence the 'extra terms' just add up to $0$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 14 '18 at 6:20

























                answered Dec 13 '18 at 8:22









                Kavi Rama MurthyKavi Rama Murthy

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                72.5k53170






























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