How do you visualize the complex function (1+i)i and its multivalued nature?

[magnifik]

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

Thanks

 

[SammyS]

Put it back into a + bi form.

 

[jackmell]

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:

$$f(z)=z^i$$

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.

 

[magnifik]

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:

$$f(z)=z^i$$

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.

what do you mean by "the sheets"? how do i know where those are located

 

[SammyS]

For n=0, this is approximately 0.428829 + 0.154872 i

 

[jackmell]

what do you mean by "the sheets"? how do i know where those are located

I skipped a lot. You can write:
$$\begin{equation}<br /> \begin{align*}<br /> z^i&=e^{i\log(z)}=e^{i(\ln|z|+i\arg(z))}\\<br /> &=e^{-\arg(z)+i\ln|z|}<br /> \end{align*}<br /> \label{eq:}<br /> \end{equation}$$
It's the $\arg(z)$ function that's multivalued and that function causes the plot to twist over itself. It's easy to draw $\arg(z)$. It's just the function $f(r,\theta)=\theta$ and the Mathematica code is simply:

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 $e^{-\arg(z)}$ represents the real part of the function $f(z)=z^i$. However, it's difficult to draw that function because it's exponential. But it's qualitatively the same as just $\arg(z)$. 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 $e^{-\arg(z)}$. Here's the complete code to show that:

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}}]