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
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          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





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          1 Answer
          1






          active

          oldest

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          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




















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