New answers tagged complex
0
I don't think you need to define upvalues for f nor for your own myConjugate function. Using your definitions:
ClearAll[F,f];
f[k_] := F[r] Exp[-I k r]
all you need to do is tell Conjugate to distribute over addition and then Refine it by letting Mathematica know which variables are real. So, say you have an expression:
Conjugate[f[k1] + f[k2] - ...
2
Assuming you want to express the solution with Re[z] and Im[z], here is a solution :
ComplexExpand[
Arg[((1 - E^((I \[Pi] (1 - \[Alpha]))/(\[Beta] - \[Alpha])) z)/(1 -
E^(-((I \[Pi] (1 - \[Alpha]))/(\[Beta] - \[Alpha]))) z))], {z},
TargetFunctions -> {Re, Im}] //Simplify
$\text{ArcTan}\left[\frac{1-\text{Cos}\left[\frac{\pi (-2+\alpha ...
2
My idea is first rewriting the complex function to $a+ib$ form then extract the real and imaginary part. So ComplexExpand and Cases is needed.
Here is my code
expr = (1 - E^((I \[Pi] (1 - \[Alpha]))/(-\[Alpha] + \[Beta])) z)/(
1 - E^(-((I \[Pi] (1 - \[Alpha]))/(-\[Alpha] + \[Beta]))) z);
newexpr = expr /. z -> (r E^(I \[Theta]));
{realPart, imPart} = ...
2
Here's one path... define your function and translate to trig form:
f[x_, z_] := ExpToTrig[((1 -
E^((I \[Pi] (1 - \[Alpha]))/(\[Beta] - \[Alpha])) z)/(1 -
E^(-((I \[Pi] (1 - \[Alpha]))/(\[Beta] - \[Alpha]))) z))]//.
(\[Pi] (1 - \[Alpha]))/(-\[Alpha] + \[Beta]) -> x
using a replacement of a new real variable x for the more complicated ...
6
The numbers you have, like 0.0161028 - 0.0119647 I are floating point (machine precision) numbers. Whenever you do calculations with floating point numbers, you will get roundoff errors. Note that you appear to have 7 or 8 digits of accuracy in the numbers. The "error" in the imaginary part is in the 18th decimal place. This is roundoff error. See this for a ...
2
You can use reciprocal to avoid $\infty$ points:
ParametricPlot[Evaluate[
ComplexExpand[
Through[{Re, Im}[(x + y I + I)/(x + y I - I) /. {x -> x^#1, y -> y^#2}]]] & @@@
Flatten[Outer[List, {1, -1}, {1, -1}], 1]
], {x, 0, 1}, {y, 0, 1}, PlotRange -> {{-3, 3}, {-.5, 4}}] // Quiet
1
Since a complex function is involved, it seems natural to use plotting functions that treat complex objects directly, without having to overtly separate them into real and imaginary parts. David Park's Presentations add-on (http://home.comcast.net/~djmpark/DrawGraphicsPage.html) allows this:
<< Presentations`
With[{f = Function[z, (z + I)/(z - I)],
...
4
I like all the existing answers because they reveal the nature of this map--but they are all deceiving. The problem is that they miss obvious and important parts of the region that cannot easily be drawn parametrically because they are associated with extremely large values of $x$ and $y$.
Here is a simple solution that is natural in the sense that it ...
2
Another approach:
Simplify[ExpToTrig @ Re[E^(I x)], x ∈ Reals]
Cos[x]
3
Try!
(1/2)*(Exp[x*I] + Conjugate[Exp[x*I]])// ComplexExpand
Cos[x]
Also note that TraditionalForm is a formatting tool mainly for displaying the formulas in a nice manner. After you have applied TraditionalForm on an expression other Mathematica built in functions (e.g FullSimplify) may have problem to deal with the expression as an input. Hence ...
3
Maybe you could expand the Vin to Fourier series to "normalize" it.
For example there are three of them kind:
VinSet = {10 Cos[1000 t - π/2], 9 Cos[400 t + π/4], Cos[t + 3.45]};
coeffSet = FourierCoefficient[# /. Times[ω_?NumericQ t] :> t, t, 1] & /@ VinSet
$\left\{-5 i,\frac{9 \sqrt[4]{-1}}{2}, -0.476409-0.151771 i\right \}$
{2 Abs[#], ...
11
ImageForwardTransformation[] is the function you want here. To give a concrete example, here's how an image might be transformed by the complex mapping $w=z^3$:
img = ExampleData[{"TestImage", "Mandrill"}];
imgc = ImageForwardTransformation[img, Through[{Re, Im}[(#[[1]] + I #[[2]])^3]] &,
Background -> 1,
...
1
As for your second question, here's one way that is very similar to what we'd do by hand. Use the exponential form and then identify the phase and magnitude.
Clear[A, p, t]
Vin = 10 Exp[-Pi/2]*Exp[1000 t I];
{A, p} = Replace[Vin, A_ Exp[p_] -> {A Exp[Re[p]], Im[p]}]
A now holds the amplitude and p holds the phase. If you have the trigonometric form you ...
2
If you specify the angle as a real number (rather than an exact integer), it does not do the transformation to Sin. For instance
Vin = 10 Cos[1000 t - Pi/2.0]
and
Vin = 10 Cos[1000 t - 90.0 Degree]
both do what you ask.
0
This is the spherical harmonic:
SphericalHarmonicY[1, 1, \[Theta], \[Phi]]
It returns this:
-(1/2) E^(I \[Phi]) Sqrt[3/(2 \[Pi])] Sin[\[Theta]]
And this is its complex conjugate:
SphericalHarmonicY[1, 1, \[Theta], \[Phi]] /. I -> -I
returning this:
-(1/2) E^(-I \[Phi]) Sqrt[3/(2 \[Pi])] Sin[\[Theta]]
as it can be expected.
1
Almost always in such situations, ComplexExpand is your friend:
Conjugate[SphericalHarmonicY[1, 1, \[Theta], \[Phi]]] // ComplexExpand
(* -(Sqrt[3/(2*Pi)]*Cos[\[Phi]]*Sin[\[Theta]])/2 +
(I/2)*Sqrt[3/(2*Pi)]*Sin[\[Theta]]*Sin[\[Phi]] *)
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