Items tagged with imaginary

I solve a set of equations in this way and I have three set of answers ,but I don`t know wich one is true.

and I have another question ,how can I assume v[0] like a constant?


alpha[2]:= 2.727272728*10^5: alpha[4]:= 3738.685337: alpha[6]:= -30.18675539: alpha[7] := -4.116375735*10^6: alpha[8] := 1.859504132*10^10: alpha[9]:= 2.489142857*10^(-12):

l10:=(alpha[7]*v[0]^2+1)*gamma[i*n]^4+(-alpha[4]*beta[n]^2+alpha[8]*v[0]^2-alpha[9])*gamma[i*n]^2+(2*I)*gamma[i*n]*alpha[2]*beta[n]*v[0]+(2*I)*gamma[i*n]^3*alpha[6]*beta[n]*v[0]-beta[n]^2 = 0:

l11 := subs(i = 1, l10);

l12 := subs(i = 2, l10);

l13 := subs(i = 3, l10);

l14 := subs(i = 4, l10);

l15 := (exp(I*(gamma[n]+gamma[2*n]))+exp(I*(gamma[3*n]+gamma[4*n])))*(gamma[3*n]^2-gamma[4*n]^2)*(gamma[n]^2-gamma[2*n]^2)+(exp(I*(gamma[n]+gamma[4*n]))+exp(I*(gamma[2*n]+gamma[3*n])))*(gamma[2*n]^2-gamma[3*n]^2)*(gamma[n]^2-gamma[4*n]^2)+(exp(I*(gamma[2*n]+gamma[4*n]))+exp(I*(gamma[n]+gamma[3*n])))*(gamma[2*n]^2-gamma[4*n]^2)*(-gamma[n]^2+gamma[3*n]^2) = 0;

l1 := combine(expand(evalc(l15)), trig):

l2 := combine(expand(evalc(Re(l15))), trig):

l3 := combine(expand(evalc(Im(l15))), trig): v[0] := 1; 1

fsolve({l1, l11, l12, l13, l14}, {beta[n], gamma[n], gamma[2*n], gamma[3*n], gamma[4*n]}):

fsolve({l11, l12, l13, l14, l2}):

solve({l11, l12, l13, l14, l3}):





I have a function:

v0(t) = -g*t-vs*ln(r*t-m0)+vs*ln(-m0)

This function should be equal to 300, but when using fsolve we get a negative real part, and a very small imaginary part:

We have already made a plot of the function, and from that we see that the t should be about 65 when v0(t)= 300.


What are we doing wrong?

Is there a way to extract the data from the graph?

hi all

i have a set of complex numerics, so:

1- i want the numeric with least valence(potency) in imagin particle,

2- i want print the real particle of this numeric.

for example:

A:= .5464691235-.4473247264*I, -.4563184747+1.*10^(-14)*I, .5464691235+.4473247264*I

i want print: -.4563184747


plz help



I know this has been dealt with before here, but I have forgotten the proper way to trim a small imaginary round-off from a result. I cannot locate the proper answer in MaplePrimes; and the Mapledocs are either quiet about it or it is hidden in a difficult-to-find place.

What I want is to ignore ny imaginary part below a certain threshold. Ideally, it takes the threshold in relation to the real part but I am not particular about this.

Thanks in advance,

Mac Dude


Im working on a assignment on maple. I have an equation of motion that looks like this:  v(t):= -g*t-vs*ln(r*t-m)+vs*ln(-m)
Im supposed to use this equation and solve it for t an later integrate it. Since the constant inside the ln is negative I end up with a annoying imaginary part. Is there any way to covert this equation so that the ln disappear, so that I can get a result whitout a imaginary part?

If we a complex number in Maple, like for example:



How can we make maple rewrite it like this?



I tried using the comands Re(%) and Im(%) but Re just gives the whole expression again and Im gives 0.

Hello everybody,

I want to find all of roots of the complex variables functions in two ways.

(1) find the value which can make the function equals 0

(2) find the real value and imaginary value which make real part and imaginary part of function equal 0

(I know answers of these two case is not equal completely.)


The function is a non-linear function, including sin, cos and Bessel function, such as:


And, I used Analytic and fsolve to do case (1) and (2), but failure. The follow result is how I tried to find the real value answer:


It seems that both of two commands can only find some of roots. 

How to find all of roots of these cases? The related .mw file is attached.


Thanks a lot.


Hi , everyone who love Maple and dsolve command, 

my ODE is :

sys_ode := diff(d11(m), m) = -(3*sin(m)^2-1)*d31(m)/a^(3/2)+(-3*cos(m)*sin(m)/a^(3/2))*d41(m), diff(d21(m), m) = (-3*cos(m)*sin(m)/a^(3/2))*d31(m)-(3*cos(m)^2-1)*d41(m)/a^(3/2), diff(d31(m), m) = -a^(3/2)*d11(m), diff(d41(m), m) = -a^(3/2)*d21(m)

using " dsolve([sys_ode]) " command could get the solution easily, and the solution contains "I" (imaginary domain).

However, when we substitute the solution into the ODE "sys_ode", find not correct !

we use the following command to check the solution :

 simplify(  -diff(d11(m), m) -(3*sin(m)^2-1)*d31(m)/a^(3/2)+(-3*cos(m)*sin(m)/a^(3/2))*d41(m)  )

the upper expression is supposed to be zero, but not ! Is it a bug in Maple dsolve ?


I'm trying to solve the equation of a form like,

diff(eta(tau), tau, tau)+(8/(4*tau^2+1)-32/(4*tau^2+1)^2)*eta(tau) = 0,

when I'm doing solve DE, I get a solution as:=

eta(tau) = _C1*sqrt(4*tau^2+1)*LegendreP((1/2*I)*sqrt(7)-1/2, I*sqrt(7), (2*I)*tau)+_C2*sqrt(4*tau^2+1)*LegendreQ((1/2*I)*sqrt(7)-1/2, I*sqrt(7), (2*I)*tau

which is combination of Legendre Polynomials with imaginary arguments,May I change this form,

How can I plot this solution on real plane, as this is imaginary,

Is the only option remaining NUMERIC PLOT??

Recently I was presented with some code that had a seq command that used i as the index variable. While this code had been used in a number of applicaitons, it was now giving trouble.

seq( i^2, i=1..10 );
Error, illegal use of an object as a name

After a stroke of fortune it was discovered that the problem was that Maple's imaginary unit had been changed from I to i. Given that this change was made, the error message now made sense.

How do I use the letter I as a variable, not an imaginary number (the square root of -1) ?

I am trying to solve two differential equations numericaly:

I am trying to get the real and imaginary part of a complex expression but it appears that the evalc(Im(F)) does not work. My guess is that i have to manipulate the expression before i use the evalc, but the simplify command for example did not give a result either. My funstion is F:=1+2*zs*I-w^2/ws^2-w^2/ws^2*ks*(1+2*zs*I)*(K3*(1+2*z3*I)-H*K2(1+2*z2*I)+H*(H*K1*(1+2*z1*I)-K2*(1+2*z2*I)))/(K1*(1+2*z1*I)*K3*(1+2*z3*I)-K2^2*(1+2*z2*I)^2);

As you see I have imaginary I...




It is pretty easy to get the principal value of the logarithm of the imaginary unit, say. Is it just as easy to get all values of the logarithm of the imaginary unit?

Forgive me if this is a duplicate, but I couldn't find a similar questions.


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