Items tagged with complex complex Tagged Items Feed

how to plot ln(sqrt(x-12)/(-x^(2)+15x)

Whenever I try to plot I get 

"Warning, unable to evaluate the function to numeric values in the region; complex values were detected".

HELP PLEASE

Hi there, 

I'm trying to find a way to find the Preimage for a single point (or a small set) of a complex rational function. I've tried with the command RationalMapPreimage, which sounds like the thing to try, but I can't get my head around how this command works?

(For context, I'm trying to find the preimages of a the fixed/critical/postcritial points of a function something like f(z)=(z2-1)/(z2+1))

Many Thanks

I have a vector, it is a zero vector apparently. I don't know what opperation lead to it being a zero vector. Any time I try to reduce this thing, I mess it up. I can't reduce it manually.

a=-((4*I)*sqrt(3)+4*sqrt(3)-5-7*I)/((4*I)*sqrt(3)+4*sqrt(3)-7-9*I)

b=-((390*I)*sqrt(3)+30*sqrt(3)-52-675*I)/(-3-4*I+2*sqrt(3)+(2*I)*sqrt(3))^3

c=-(1/2)*sqrt(2-2*I)*((6*I)*sqrt(3)+6*sqrt(3)-11-10*I)*(-2+sqrt(3))*sqrt(-2-2*I)*((2*I)*sqrt(3)+2*sqrt(3)-5-2*I)/(-3-4*I+2*sqrt(3)+(2*I)*sqrt(3))^3

 

<a/sqrt(a^2+b^2+c^2),b/sqrt(a^2+b^2+c^2),c/sqrt(a^2+b^2+c^2)>

 

The simplify command will reduce this bugger no problem. The moment I try rationalizing the denominator or anything like that I end up buggering the whole thing up. Maybe I should do a distance on two of the ordinates independently first. Anyway, I don't know what's up. 

Hello everyone! I got some trouble in process a list. Hope you can help:

Assume i got a list like this:

 

{{k = k, l = RootOf(_Z^2+_Z*k+k^2-1), o = -k-RootOf(_Z^2+_Z*k+k^2-1)}, {k = k, l = RootOf(_Z^2+_Z*k+k^2+1), o = -k-RootOf(_Z^2+_Z*k+k^2+1)}, {k = 0, l = 1, o = -1}, {k = 0, l = -1, o = 1}, {k = 1, l = 0, o = -1}, {k = 1, l = -1, o = 0}, {k = -1, l = 0, o = 1}, {k = -1, l = 1, o = 0}, {k = RootOf(_Z^2+1), l = 0, o = -RootOf(_Z^2+1)}, {k = RootOf(_Z^2+1), l = -RootOf(_Z^2+1), o = 0}}

 

Now all i want is remove Complex and RootOf from this list, how can i do that?

Thank for your reading adn your help!

Hello!

 

If I type this into a CAS calculator from Texas, I get the right result for Uc wich is 545.2.

solve(Uc&angle;-90=400&angle;X-150&angle;0+j*174.4,Uc,X)

This is used in AC circuits.

However, in Maple I'm using polar coordinates, so Im using this preamble:

phasor := proc (r, theta) options operator, arrow; r*cos(convert(theta*degrees, radians))+I*r*sin(convert(theta*degrees, radians)) end proc

This basically means that I can write phasor(value,angle in degrees) and get the lenth and angle of a vector on the complex plane.

Basically I want to know how I can make Maple solve this:

evalf(solve({phasor(U__C, -90) = phasor(400, X)-phasor(150, 0)+phasor(174.4, 90)}, {Uc, X})) Where I'm just interested in U_C , X doesn't matter.

I can't figure out how to do this with my current preamble. Does anyone know how to make this work?

Hello,

I would like to know how to order a sequence of number from smallest to largest. This is if I have both real and imaginary numbers. Any help would be rgeatly appreciated! Thank you in advance.

Kind regards,

Gamiba Man

There are the complexes (C), quaternions (H or Q), octionions (O), sedenions (S) and the pathions (P). I have found the multiplication tables of them, although according to signs (+ or -) there differents at pathions. The important question is that How can I multiply two bases, i_n and i_m of higher dimensions, like in the routions or in the voudions?

Should I xor the indexes of the bases? Like this way: i_1 * i_2 = i_(1^2) = i_3

What is about the signs?

Hi,

 

I am trying to find the roots of Hankel function H1(2, z)?

 

j := 1;
for i from 0 to 10 do z0 := i*step+z_min; x[j] := fsolve(f = 0, z = z0, complex) end do;


for p to 10 do for j from p+1 to 9 do if `and`(Re(x[j])-Re(x[p]) < 0.1e-4, Im(x[j])-Im(x[p]) < 0.1e-4) then for i from j to 10 do x[i] := x[i+1] end do; p := p-1; break end if end do end do;

 

the first of these two processes works fine however the second does not. The second on is to get rid of same value solutions! I am not sure if I have missed anything and also is there a way to determine a max value in the complex domain and use it in a for loop?

 

 

any help would be great 

How does one 3D plot the simple complex exponential e^I2pift or cos(2pift)+I*sin(2pift)  where f is freqency and t is time.  It shoud display a spiral aroung the time axis.

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

(-6.328281880*10^(-49)-1.071713312*10^(-47)*I)*exp(-5996.664400+2.000000000*10^14*del+(1.547600000*10^16*I)*del)*(erf(11.+77.38000000*I+1.000000000*10^14*del)-1.*erf(-4.+77.38000000*I+1.000000000*10^14*del))

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 ...

 

 

Regards,

Erik

 

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));

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

ln(rho)+Complex(1)*theta;=ln(rho)+Complex(1)*arctan(y,x);

and

z^a=exp(a*ln(z));

and

"Because of

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

and

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.

 

taro

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

http://prntscr.com/8abmta

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.

 

    

restart:

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

simplify(%);

Thanks

1 2 3 4 5 6 7 Last Page 1 of 27