What's the Laplace operator in a complex plane $z=x+iy$?












0












$begingroup$


Firstly, I don't know complex analysis.



Can the Laplace operator in a complex plane $z=x+iy$ be expressed as
$$nabla=frac{partial}{partial x}+ifrac{partial}{partial y}$$
or
$$nabla=frac{partial}{partial x}+frac{partial}{ipartial y}=frac{partial}{partial x}-ifrac{partial}{partial y}$$
Which is right? or not one?



I originally wanted to verify the power series $V=C_n(x+iy)^n$ in complex plane $z=x+iy$ is exactly a solution for a Laplacian equation
$$nabla^2V=0$$. How to do that?






complex-analysis harmonic-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Firstly, I don't know complex analysis.



    Can the Laplace operator in a complex plane $z=x+iy$ be expressed as
    $$nabla=frac{partial}{partial x}+ifrac{partial}{partial y}$$
    or
    $$nabla=frac{partial}{partial x}+frac{partial}{ipartial y}=frac{partial}{partial x}-ifrac{partial}{partial y}$$
    Which is right? or not one?



    I originally wanted to verify the power series $V=C_n(x+iy)^n$ in complex plane $z=x+iy$ is exactly a solution for a Laplacian equation
    $$nabla^2V=0$$. How to do that?






    complex-analysis harmonic-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Firstly, I don't know complex analysis.



      Can the Laplace operator in a complex plane $z=x+iy$ be expressed as
      $$nabla=frac{partial}{partial x}+ifrac{partial}{partial y}$$
      or
      $$nabla=frac{partial}{partial x}+frac{partial}{ipartial y}=frac{partial}{partial x}-ifrac{partial}{partial y}$$
      Which is right? or not one?



      I originally wanted to verify the power series $V=C_n(x+iy)^n$ in complex plane $z=x+iy$ is exactly a solution for a Laplacian equation
      $$nabla^2V=0$$. How to do that?






      complex-analysis harmonic-functions







      share|cite|improve this question











      $endgroup$




      Firstly, I don't know complex analysis.



      Can the Laplace operator in a complex plane $z=x+iy$ be expressed as
      $$nabla=frac{partial}{partial x}+ifrac{partial}{partial y}$$
      or
      $$nabla=frac{partial}{partial x}+frac{partial}{ipartial y}=frac{partial}{partial x}-ifrac{partial}{partial y}$$
      Which is right? or not one?



      I originally wanted to verify the power series $V=C_n(x+iy)^n$ in complex plane $z=x+iy$ is exactly a solution for a Laplacian equation
      $$nabla^2V=0$$. How to do that?






      complex-analysis harmonic-functions




      complex-analysis harmonic-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 3:05









      hardmath

      28.9k95297




      28.9k95297










      asked Mar 8 '17 at 10:30









      FeynmanFeynman

      1667




      1667






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          You can identify $mathbb{R}^2 cong mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.

          Then, the following relationship holds:

          If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $mathbb{R}^2 to mathbb{R}$.

          There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):

          $$
          Delta u=frac{1}{4}frac{partial^2 u}{partial z partial bar{z}}
          $$

          Now, to look at your function we have that and using that $i^2=-1$:
          $$
          frac{partial u}{partial z}=frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \
          C_nn(x+iy)^{n-1}
          $$
          $$
          frac{partial^2 u}{partial z partial bar{z}}=frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0
          $$
          if your n=1, then the second step is not needed. If $n geq 2$, the calculation is fine.
          If that was not the answer you were looking for, dont hesitate to ask!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Tanks! But what's the $nabla$?
            $endgroup$
            – Feynman
            Mar 8 '17 at 11:19










          • $begingroup$
            There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
            $endgroup$
            – F. Conrad
            Mar 8 '17 at 11:31










          • $begingroup$
            Yes. It is very useful
            $endgroup$
            – Feynman
            Mar 9 '17 at 0:05













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2177365%2fwhats-the-laplace-operator-in-a-complex-plane-z-xiy%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          You can identify $mathbb{R}^2 cong mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.

          Then, the following relationship holds:

          If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $mathbb{R}^2 to mathbb{R}$.

          There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):

          $$
          Delta u=frac{1}{4}frac{partial^2 u}{partial z partial bar{z}}
          $$

          Now, to look at your function we have that and using that $i^2=-1$:
          $$
          frac{partial u}{partial z}=frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \
          C_nn(x+iy)^{n-1}
          $$
          $$
          frac{partial^2 u}{partial z partial bar{z}}=frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0
          $$
          if your n=1, then the second step is not needed. If $n geq 2$, the calculation is fine.
          If that was not the answer you were looking for, dont hesitate to ask!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Tanks! But what's the $nabla$?
            $endgroup$
            – Feynman
            Mar 8 '17 at 11:19










          • $begingroup$
            There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
            $endgroup$
            – F. Conrad
            Mar 8 '17 at 11:31










          • $begingroup$
            Yes. It is very useful
            $endgroup$
            – Feynman
            Mar 9 '17 at 0:05


















          0












          $begingroup$

          You can identify $mathbb{R}^2 cong mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.

          Then, the following relationship holds:

          If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $mathbb{R}^2 to mathbb{R}$.

          There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):

          $$
          Delta u=frac{1}{4}frac{partial^2 u}{partial z partial bar{z}}
          $$

          Now, to look at your function we have that and using that $i^2=-1$:
          $$
          frac{partial u}{partial z}=frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \
          C_nn(x+iy)^{n-1}
          $$
          $$
          frac{partial^2 u}{partial z partial bar{z}}=frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0
          $$
          if your n=1, then the second step is not needed. If $n geq 2$, the calculation is fine.
          If that was not the answer you were looking for, dont hesitate to ask!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Tanks! But what's the $nabla$?
            $endgroup$
            – Feynman
            Mar 8 '17 at 11:19










          • $begingroup$
            There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
            $endgroup$
            – F. Conrad
            Mar 8 '17 at 11:31










          • $begingroup$
            Yes. It is very useful
            $endgroup$
            – Feynman
            Mar 9 '17 at 0:05
















          0












          0








          0





          $begingroup$

          You can identify $mathbb{R}^2 cong mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.

          Then, the following relationship holds:

          If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $mathbb{R}^2 to mathbb{R}$.

          There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):

          $$
          Delta u=frac{1}{4}frac{partial^2 u}{partial z partial bar{z}}
          $$

          Now, to look at your function we have that and using that $i^2=-1$:
          $$
          frac{partial u}{partial z}=frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \
          C_nn(x+iy)^{n-1}
          $$
          $$
          frac{partial^2 u}{partial z partial bar{z}}=frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0
          $$
          if your n=1, then the second step is not needed. If $n geq 2$, the calculation is fine.
          If that was not the answer you were looking for, dont hesitate to ask!






          share|cite|improve this answer











          $endgroup$



          You can identify $mathbb{R}^2 cong mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.

          Then, the following relationship holds:

          If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $mathbb{R}^2 to mathbb{R}$.

          There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):

          $$
          Delta u=frac{1}{4}frac{partial^2 u}{partial z partial bar{z}}
          $$

          Now, to look at your function we have that and using that $i^2=-1$:
          $$
          frac{partial u}{partial z}=frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \
          C_nn(x+iy)^{n-1}
          $$
          $$
          frac{partial^2 u}{partial z partial bar{z}}=frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0
          $$
          if your n=1, then the second step is not needed. If $n geq 2$, the calculation is fine.
          If that was not the answer you were looking for, dont hesitate to ask!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 8 '17 at 11:12

























          answered Mar 8 '17 at 11:07









          F. ConradF. Conrad

          1,243412




          1,243412












          • $begingroup$
            Tanks! But what's the $nabla$?
            $endgroup$
            – Feynman
            Mar 8 '17 at 11:19










          • $begingroup$
            There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
            $endgroup$
            – F. Conrad
            Mar 8 '17 at 11:31










          • $begingroup$
            Yes. It is very useful
            $endgroup$
            – Feynman
            Mar 9 '17 at 0:05




















          • $begingroup$
            Tanks! But what's the $nabla$?
            $endgroup$
            – Feynman
            Mar 8 '17 at 11:19










          • $begingroup$
            There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
            $endgroup$
            – F. Conrad
            Mar 8 '17 at 11:31










          • $begingroup$
            Yes. It is very useful
            $endgroup$
            – Feynman
            Mar 9 '17 at 0:05


















          $begingroup$
          Tanks! But what's the $nabla$?
          $endgroup$
          – Feynman
          Mar 8 '17 at 11:19




          $begingroup$
          Tanks! But what's the $nabla$?
          $endgroup$
          – Feynman
          Mar 8 '17 at 11:19












          $begingroup$
          There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
          $endgroup$
          – F. Conrad
          Mar 8 '17 at 11:31




          $begingroup$
          There is no such thing as "the gradient". I think you are looking for the wirtinger derivative.
          $endgroup$
          – F. Conrad
          Mar 8 '17 at 11:31












          $begingroup$
          Yes. It is very useful
          $endgroup$
          – Feynman
          Mar 9 '17 at 0:05






          $begingroup$
          Yes. It is very useful
          $endgroup$
          – Feynman
          Mar 9 '17 at 0:05




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2177365%2fwhats-the-laplace-operator-in-a-complex-plane-z-xiy%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          Dobbiaco

          Image
          Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
          Read more