# Plot complex function

1. Oct 6, 2011

### magnifik

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. Oct 6, 2011

### SammyS

Staff Emeritus
Put it back into a + bi form.

3. Oct 7, 2011

### 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.

4. Oct 7, 2011

### magnifik

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

5. Oct 7, 2011

### SammyS

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

6. Oct 8, 2011

### jackmell

I skipped a lot. You can write:
\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:}
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:

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 $e^{-\arg(z)}$ 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:

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:

• ###### my arg function.jpg
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Last edited: Oct 8, 2011