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how to plot exp(f)*erf(g) where f and g are complex function


del from -100e-15 to 100e-15

Dear Maple users

An engineering student asked me how Maple is handling complex numbers in polar form. He told me that his fellow students are using another CAS, whereas he himself prefer Maple. When making calculations with AC currents having different phases the other students were using the easy notation depicted in the first line on the picture below. Obviously here the angle (argument) is measured in degrees. I tried to perform the same calculations in Maple, but found it to require a very heavy notation: the other three lines on the picture. Now my question is: Does it really have to be that messy, or maybe there are some package, which will accomplish the task in a more neat way? I mean it is a rather common operation in the engineering sciences.

NB! Of couse one can argue about the educational value of using the notation of the other CAS! From that viewpoint they will probably not learn anything ...






Hello people in mapleprimes,

I have a question about why what is shown by maple by simplify(-8)^(1/3) is 1+ Complex(1)* (3)^(1/2)?

Solutions of x^3=-8 are -2, 1+Complex(1)*(3)^(1/2) and 1- Complex(1)*(3)^(1/2). And, as for the last one, 1- Complex(1)*(3)^(1/2), it is 

the conjugate of the second, so it might not need to be written, because of it being easily seen so.

Is it the same reason why just -2 is not shown as the result of simplify((-8)^(1/3))?

PS. I know the instruction to use surd in such a case.

the reason I asked this question is this:

I am reading Essential Maple, where

ln(z) = ln(rho*exp(Complex(1)*theta));






"Because of

exp(w*Pi*Complex(1)*k)=cos(2*Pi*k) + Complex(1)*sin(2*Pi*k);


cos(2*Pi*k) + Complex(1)*sin(2*Pi*k)=1;

we could equally well have chosen

ln_k z = ln(z) + 2*Pi*Complex(1)*k"

are written.

 Supposing these, there is a sentence that

"we choose k=0, and thus -Pi<=theta<=Pi to be the one (that for our canonical logarithm).

Every computer algebra language and numerical language follows this standard and takes the

complex logarithm to have its imaginary part in this range.

With this definition, (-8)^(1/3)=1 + Complex(1)*sqrt(3), and not -2. (the end of quotation)"


And, I can't understand the last sentence"With this definition", so I asked the above question.


I hope someone give an answer to the above question.


Thanks in advance.



I'm not sure why im getting a complex solution for evalf(h(-1/2)). Posted screenshot here:

The answer should be positive 6*2^(2/3) ≈ 9.52

 The computer returns

h(-1/2) =


The problem is that evalf((-1)^(1/3)) you get 0.500 + .866I

Is there no way to evaluate a second derivative of a real valued function which has a fractional exponent without receiving complex results? I don't have the time to look at each function and try to figure out what went wrong. I want to plug in any x value into a function defined for all reals and get a real result.

I tried  assume(x , 'real' ) , that did not do anything.




f:=exp(-I*Im(s)*(t)); # I is iota, s is a complex number and t is time.



Hello everybody,


my question concerns the visualizing possibilities of maple:

can maple visualize complex functions? (f(z): C->C)

If so, which possibilities do i have und what is the command for it? Maybe as a coulour diagram, as a vector field (f(x): R^2->R^2) , or as mapping of sets (e.g. curves, grids into new curves and curved lines)?




Thanks in advance




PS: I am using maple 18.

A small piece of code for fun before the weekend, inspired by some recent posts:


cycler := proc(t, k, p, m, n, T) local expr, u, v;
  u := exp(k*I*t);
  expr := exp(I*t) * (1 - u/m + I*(u^(-p))/n);
  v := 1 + abs(1/n) + abs(1/m);
  plots:-complexplot( expr, t = 0 .. T, axes = none,
                      view = [-v .. v, -v .. v] );
end proc:

cycler(t, 5, 3, 2, 3, 2*Pi);

And that can be made into a small application (needs Maple 18 or 2015),

Explore( cycler(t, k, p, m, n, T),
         parameters = [ k = -10 .. 10, p = -10 .. 10,
                        m = -10.0 .. 10.0, n = -10.0 .. 10.0,
                        [ T = 0 .. 2*Pi, animate ] ],
         initialvalues = [ k = 5, p = 3, m = 2, n = 3, T = 2*Pi ],
         placement = left, animate = false, numframes = 100 );

The animation of parameter T from 0 to 2*Pi can also be fun if the view option is removed/disabled in the call to complexplot. (That could even be made a choice, by adding an additional argument for calling procedure cycler and an accompanying checkbox parameter on the exploration.)


I want to solve the int((y4+by2+c)-1/2,y)-x and find y=h(x), where b and c are constants s.t. c>b2/4. Maple gives me complex Jacobi elliptic function as a result. But I am not sure that this integral has complex value. Am I doing something wrong or the result is really a complex valued function? Thanks.

Indeed my main question is: Plot y=y(u) where we have these two relations: int((y4+by2+c)-1/2,y)=x and find y=h(x). Then evaluate int((h(x)-B)-1/2,x)=u and find x=g(u). By using these relations plot y=y(u). :)

Here B is an arbitrary constant, but if necessary we can define a value for it. All the variables and constants are real.

I hope I manage to express myself. Thanks again.



I have a perfectly working when all parameters are known (figure 1), however I want to perform a sensitivity analysis by derivating the code if one parameter is unknown. Because of multiple possible answers and because of the complexity of the formula, I cannot run this script and get solutions. Any ideas how I can this calculation lighter so it is able to run? Values should be real and positive (so 1 or 2 solutions are the only one I'm interested in)

Any ideas, how I can make this code runnable? (file is below)

I'm stuck on this for a while now :/ So I hope someone will be able to help me

Many thanks in advance!l

Figure 1: [URL=][IMG][/IMG][/URL]


Figure 2: [URL=][IMG][/IMG][/URL]

I have the following characteristic equation by use of maple. How do I find a condition on x, that will return real eigenvalues and complex eigenvalues?




If you have several results, e.g. one being real, the others complex. How can you choose?

G := 6.673*10^(-11):

M := 1.9891*10^30:

AU := 1.4959787066*10^11:

v := proc (R) options operator, arrow; 2332800000/(Pi*(R^3*AU^3/(G*M))^.5) end proc;

proc (R) options operator, arrow; 2332800000/(Pi*(R^3*AU^3/(G*M))^.5) end proc


a := solve(.601 = 60*(360*60)/(2*Pi*((R*AU)^3/(G*M))^.5)*3600, R);

39.26171595, -19.63085797+34.00164341*I, -19.63085797-34.00164341*I


op(1, a)

Error, invalid input: op expects 1 or 2 arguments, but received 4




Hi everyone,

at the suggestion of Carl I am making my question a post.


Newbies often get fascinated with the power tower: x^x^x^...

The generalized power tower z^z^z^... is a special case of the Euler sequence:

z_{n+1}=c^z_n, for c\in C.

Like the Mandelbrot set: z_{n+1}=z^2-c, the power tower, often called infinite exponential, also has a general periodic map. It is included below and is taken from Daniel Geisler's site:

Shel and Thron proved that the infinite exponential conveges whenever c belongs to the red region, called today Shell-Thron region.

Definition of Julia Sets for the iterated exponential map:

Also like the Mandelbrot set, the infinite exponential admits Juila Sets. The Julia Sets of the infinite exponential however, are defined differently from the Julia Sets of the Mandelbrot set. They are defined to be  the closure of the periodic repellers of the Euler sequence . They are Cantor Bouquets.  Geisler's colored map then is a general map of how the corresponding Julia set behaves roughly, with c taken from the map. 

We can then introduce small cuts which go from the interior of the Shell-Thron region to the exterior, crossing at various angles, and this will tells us how the infinite exponential evolves. Generally speaking, each time one crosses the Shell-Thron boundary, one wittnesses what's called a Knaster explosion, wherein the exponential explodes into p subregions, where p is the pre-period of the multiplier.

When the parameter c exits the Shell-Thron region at angles of 2*Pi and Pi from the real axis (cuts right and left, p=1, 2), the infinite exponential either transitions from converging to a single feature to exploding into multiple indecomposable contiua (p=1), or it breaks into a period 2 bifurcation (p=2), which itself, also may explode into continua.

When it exits at angles 2*Pi/p, where p>2 is the preperiod of the multiplier, then the infinite exponential evolves from converging to a single feature, to exploding into  p major regions, called Fatou regions, each one having its own attractor, displaying a p-furcation.

In all cases,  the Knaster explosions may introduce the presence of indecomposable continua, as some Fatou regions end up covering entire parts of the complex plane after each transition. In the animations, Knaster explosions occur whenever the background is red. There may be more than one explosion present in the evolution of the power tower. 

Cantor Bouquets are strange creatures. They are essentially quantum sets, and no point of them is actually visible. The probability a point is visible varies directly with the area of the corresponding bouquet "finger" which is rendered in the area of interest. Devaney uses these "fingers"  to obtain "iteneraries" of the iterated exponential map.

Points "close"  to the fingers-hairs o the Cantor Bouquets escape towards complex infinity at (final) angles 2*Pi/p.

The "hairs" of a Cantor Bouquet are C^\infty curves, hence they can be termed easthetically pretty. Only hairs from the main Cantor bouquet for c=e^{1/e} are globally convex/concave. Every other bouquet may contain hairs which change curvature "unpredictably".

Inlcuded files:





Code for static fractal with given parameters in 1). Parameters can be changed in the constant section (Try p=2,4 or Pi)

Code for morphing animations through cuts in the Shell-Thron region in 2). Parameters d1-d2, N

Code for zooming animations in 3). Parameters M, M1-M2,


Will show attractors of broken Fatou basins only up to pre-period p=5. No big deal though. If you want to increase the preperiod past p=5, you need to add the relavnt tests in procedure ppc(c), otherwise the new attractors are not calculated and the plot ends up red.

Colors are assigned a bit more dispersed, using ln(m). You can also use m in that place. It depends on which area you are in.

Basins of attraction and Fatou regions change appearance under different epsilons. If you want different shapes for the basins, you can try using the Manhattan metric in proc Jhf, instead of |z-lim|.

Included below are the main map of tetration by Daniel Geisler, a short excerpt of the Shell-Thron region which shows pre-periodic cuts and  6 morhing transitions, for p\in {1,2,3,4,5} and one which exits at the angle of 2 radians (p=Pi).


More than 300. To be published with my thesis. Patches, problems and code corrections in the original question page:


I got the Real and Imaginary of an expression J1 




J1mod:=simplify((Re(J1))^2+(Im(J1))^2): (I works here this amont is real)


but when I change the expression  for J1 to be



J1mod here is complex(I dont know why? it doesnt separate the real and the im )

Any comments will help



I need you help to understand this problem.


Let alpha=0.1.;

When I do :  (-0.2)^alpha I get a complex number.


I must find a real number.

What's the problem

Thanks for your idea.


Hi all.

I try to get the real part from the complex expression. But it turns out to not be the simplest result:


convert(exp(-I*k[0]*h), sin);


Maple results in:


while the simplified result should be:



I wander how to get the simplifyed result in maple. Thanks

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