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Plot complex function

  1. Oct 6, 2011 #1
    How do you plot (1+i)i, where i is the imaginary number. I decomposed it to eilog√2e-∏/4e2∏n (n = 0, +1, +2, ...) Should it be some kind of lattice? I would imagine it's discontinuous due to the n

    Thanks
     
  2. jcsd
  3. Oct 6, 2011 #2

    SammyS

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    Put it back into a + bi form.
     
  4. Oct 7, 2011 #3
    No, it's not discontinuous. The n is only used to distinguish single-valued branching. It's still one single multi-function in the complex plane with an infinitely twisted sheet. Write it as:

    [tex]f(z)=z^i[/tex]

    Now, let z=1+i and draw a vertical line above that point in the complex plane. Where ever this line hits the sheets, that's the (infinite) values of (1+i)^i.
     
  5. Oct 7, 2011 #4
    what do you mean by "the sheets"? how do i know where those are located
     
  6. Oct 7, 2011 #5

    SammyS

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    For n=0, this is approximately 0.428829 + 0.154872 i
     
  7. Oct 8, 2011 #6
    I skipped a lot. You can write:
    [tex]
    \begin{equation}
    \begin{align*}
    z^i&=e^{i\log(z)}=e^{i(\ln|z|+i\arg(z))}\\
    &=e^{-\arg(z)+i\ln|z|}
    \end{align*}
    \label{eq:}
    \end{equation}
    [/tex]
    It's the [itex]\arg(z)[/itex] function that's multivalued and that function causes the plot to twist over itself. It's easy to draw [itex]\arg(z)[/itex]. It's just the function [itex]f(r,\theta)=\theta[/itex] and the Mathematica code is simply:

    Code (Text):
    ParametricPlot3D[{Re[z], Im[z], t} /. z -> r Exp[I t], {r, 0,
      2}, {t, -4 \[Pi], 4 \[Pi]}, BoxRatios -> {1, 1, 2},
     PlotPoints -> {35, 35}]
     
    So the expression [itex]e^{-\arg(z)}[/itex] represents the real part of the function [itex]f(z)=z^i[/itex]. However, it's difficult to draw that function because it's exponential. But it's qualitatively the same as just [itex]\arg(z)[/itex]. So I'll just plot (x,y,arg(x+iy)) below: Now draw that function in the complex plane, then draw a vertical line over the point z=1+i and where ever the line intersects the plot is the multiple values of arg(1+i) and that would be conceptually the same as doing that for [itex]e^{-\arg(z)}[/itex]. Here's the complete code to show that:

    Code (Text):
    myline = Graphics3D[{Thickness[0.008], Red,
        Line[{{1, 1, -20}, {1, 1, 20}}]}];
    mypoints =
      Graphics3D[{PointSize[0.05], Blue,
        Point @@ {{1, 1, #}} & /@ {\[Pi]/4, 9 \[Pi]/4,
          17 \[Pi]/4, -7 \[Pi]/4}}];

    Show[{ParametricPlot3D[{Re[z], Im[z], t} /. z -> r Exp[I t], {r, 0,
        2}, {t, -4 \[Pi], 4 \[Pi]}, BoxRatios -> {1, 1, 2},
       PlotPoints -> {35, 35}], myline, mypoints},
     PlotRange -> {{-2, 2}, {-2, 2}, {-10, 10}}]
     
     

    Attached Files:

    Last edited: Oct 8, 2011
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