How do I visualize $\exp(z)$ as a complex mapping? How can I ensure that it does not miss any value on the complex plane as it's value (is in the best condition of Picard's theorem). Can anyone help me?
1 Answer
You could use ComplexContourPlot
for visualising the real and imaginary axes on the complex z
plane under the complex exponential mapping Exp[z]
.
Define function for complex mapping:
f[z_] := Exp[z]
Then use ComplexContourPlot
for visualising the contours:
Edit: Contours
and ContourLabels
added to ComplexContourPlot
following Michael E2 comment. This shows more clearly that the same lines are displayed in both contour plots.
{ComplexContourPlot[ReIm[z], {z, -3 - 3 I, 3 + 3 I}, PlotLabel -> z,
Contours -> {Range[-2, 2]}, ContourLabels -> All],
ComplexContourPlot[ReIm[f[z]], {z, -3 - 3 I, 3 + 3 I},
PlotLabel -> f[z], Contours -> {Range[-2, 2]},
ContourLabels -> All]} // Grid[{#}, Frame -> True] &
The result:
Also you could look into the modulus and argument of Exp[z]
. Perhaps this shows more clearly the mapping:
{ComplexContourPlot[AbsArg[z], {z, -3 - 3 I, 3 + 3 I}, PlotLabel -> z,
Contours -> {Range[-3, 3]}, ContourLabels -> All],
ComplexContourPlot[AbsArg[f[z]], {z, -3 - 3 I, 3 + 3 I},
PlotLabel -> f[z], Contours -> {Range[-3, 3]},
ContourLabels -> All]} // Grid[{#}, Frame -> True] &
Here you can see that circles in z
plane (Abs[z]
constant) are mapped into real lines Exp[x+Iy] (* Exp[x] is constant *)
, and that lines through the origin in the z
plane (Arg[z]
constant) are mapped into Exp[x+Iy] (* y is constant*)
lines.
-
1$\begingroup$ You might want to add something like
Contours -> {Range[-2, 2]}
to both plots so that the lines of both graphs correspond to each other. $\endgroup$ Commented Oct 6, 2020 at 20:11
ComplexPlot
$\endgroup$