Recover complex function from its imaginary part











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The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
$$
left{
begin{array}{c}
G_x = 2x+1 \
G_y = -2y
end{array}
right.
$$



$$
left{
begin{array}{c}
G_{xx} = 2 \
G_{yy} = -2
end{array}
right.
$$
$G_{xx} + G_{yy} = 0$, therefore the function is harmonic.



Next step is
$$
left{
begin{array}{c}
u_x = G_y = -2y \
u_y = -G_x = -2x-1
end{array}
right.
$$



Then,
$$u=int u_x dx=-2xy+h(y)+C$$
$$u_y = -2x-1$$
Therefore,
$$u = -2xy - 2x - 1$$
Finally,
$$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
But this is a wrong answer. Where is my mistake?






complex-analysis complex-numbers







share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
    $$
    left{
    begin{array}{c}
    G_x = 2x+1 \
    G_y = -2y
    end{array}
    right.
    $$



    $$
    left{
    begin{array}{c}
    G_{xx} = 2 \
    G_{yy} = -2
    end{array}
    right.
    $$
    $G_{xx} + G_{yy} = 0$, therefore the function is harmonic.



    Next step is
    $$
    left{
    begin{array}{c}
    u_x = G_y = -2y \
    u_y = -G_x = -2x-1
    end{array}
    right.
    $$



    Then,
    $$u=int u_x dx=-2xy+h(y)+C$$
    $$u_y = -2x-1$$
    Therefore,
    $$u = -2xy - 2x - 1$$
    Finally,
    $$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
    But this is a wrong answer. Where is my mistake?






    complex-analysis complex-numbers







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
      $$
      left{
      begin{array}{c}
      G_x = 2x+1 \
      G_y = -2y
      end{array}
      right.
      $$



      $$
      left{
      begin{array}{c}
      G_{xx} = 2 \
      G_{yy} = -2
      end{array}
      right.
      $$
      $G_{xx} + G_{yy} = 0$, therefore the function is harmonic.



      Next step is
      $$
      left{
      begin{array}{c}
      u_x = G_y = -2y \
      u_y = -G_x = -2x-1
      end{array}
      right.
      $$



      Then,
      $$u=int u_x dx=-2xy+h(y)+C$$
      $$u_y = -2x-1$$
      Therefore,
      $$u = -2xy - 2x - 1$$
      Finally,
      $$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
      But this is a wrong answer. Where is my mistake?






      complex-analysis complex-numbers







      share|cite|improve this question













      The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
      $$
      left{
      begin{array}{c}
      G_x = 2x+1 \
      G_y = -2y
      end{array}
      right.
      $$



      $$
      left{
      begin{array}{c}
      G_{xx} = 2 \
      G_{yy} = -2
      end{array}
      right.
      $$
      $G_{xx} + G_{yy} = 0$, therefore the function is harmonic.



      Next step is
      $$
      left{
      begin{array}{c}
      u_x = G_y = -2y \
      u_y = -G_x = -2x-1
      end{array}
      right.
      $$



      Then,
      $$u=int u_x dx=-2xy+h(y)+C$$
      $$u_y = -2x-1$$
      Therefore,
      $$u = -2xy - 2x - 1$$
      Finally,
      $$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
      But this is a wrong answer. Where is my mistake?






      complex-analysis complex-numbers




      complex-analysis complex-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 6:11









      user3132457

      876




      876






















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          $u = -2xy - 2x - 1$ is wrong !



          From $u=-2xy+h(y)+C$ we get



          $u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.






          share|cite|improve this answer





















          • Thanks, I just don't get why you put $h'(y)$ instead of -1
            – user3132457
            Nov 22 at 12:19













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          up vote
          0
          down vote



          accepted










          $u = -2xy - 2x - 1$ is wrong !



          From $u=-2xy+h(y)+C$ we get



          $u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.






          share|cite|improve this answer





















          • Thanks, I just don't get why you put $h'(y)$ instead of -1
            – user3132457
            Nov 22 at 12:19

















          up vote
          0
          down vote



          accepted










          $u = -2xy - 2x - 1$ is wrong !



          From $u=-2xy+h(y)+C$ we get



          $u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.






          share|cite|improve this answer





















          • Thanks, I just don't get why you put $h'(y)$ instead of -1
            – user3132457
            Nov 22 at 12:19















          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          $u = -2xy - 2x - 1$ is wrong !



          From $u=-2xy+h(y)+C$ we get



          $u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.






          share|cite|improve this answer












          $u = -2xy - 2x - 1$ is wrong !



          From $u=-2xy+h(y)+C$ we get



          $u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 6:21









          Fred

          43.6k1644




          43.6k1644












          • Thanks, I just don't get why you put $h'(y)$ instead of -1
            – user3132457
            Nov 22 at 12:19




















          • Thanks, I just don't get why you put $h'(y)$ instead of -1
            – user3132457
            Nov 22 at 12:19


















          Thanks, I just don't get why you put $h'(y)$ instead of -1
          – user3132457
          Nov 22 at 12:19






          Thanks, I just don't get why you put $h'(y)$ instead of -1
          – user3132457
          Nov 22 at 12:19




















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