complex number arithmetic

[sharadbags]

how do we pass a complex no. to a C++ program
the real and imaginary values are dependent on some variables which are calculated at runtime

[horace1]

how do we pass a complex no. to a C++ program
the real and imaginary values are dependent on some variables which are calculated at runtime

have a look at the C++ complex number class
http://www.eng.cam.ac. uk/help/tpl/languages/C++/complex.html
http://support.sas.com/documentation/onlinedoc/sasc/doc/cplus/z0274855.htm

[sharadbags]

I found that in the standard library
a structure has been defined for complex no. and not as class as mentioned in 2nd website
so
1) do we have to create our own class?
2)is there a tutorial that can teach how to use functions in complex.h
are there any examples of it?

[horace1]

I found that in the standard library
a structure has been defined for complex no. and not as class as mentioned in 2nd website
so
1) do we have to create our own class?
2)is there a tutorial that can teach how to use functions in complex.h
are there any examples of it?

If you have a modern compiler <complex> should be available - what compiler are you useing?
try a small program, e.g.

#include <iostream>
#include <complex>
using namespace std;
int main() {
 complex<double> c1(0.0,1.0), c2; // complex numbers with double components
 c2=pow(c1,2.0);
 cout << "The square of " << c1 << " is " << c2 << endl; 
 system("pause");
}

[RRick]

I'm not sure what you found, but Ansi C++ does support a complex datatype. The above links do have a good description of the class. There is a minor change and you should include "complex", not "complex.h" .

This means you shouldn't have to make your own complex type.

[sharadbags]

i have pgain as a variable of data type complex<double>
now i want only it's real part to be put in a double variable
how do i do that

[sharadbags]

kindly someone reply

[Ganon11]

This tutorial should give you the answers you need.

Alternatively, you could always create your own class called complex to perform calculations. All it needs is two double variables (real and imaginary) and some operator overriding.

[horace1]

kindly someone reply

have a look at this program which determines the roots of a cubic equation
a*x^3 + b*x^2 + c*x + d =0

#include <iostream>
#include <complex>
#include <iomanip>
#include <math.h>

using namespace std;

// check root value
complex<double> check(double a, double b, double c, double d, complex<double> x)
{
    return  a*x*x*x + b*x*x + c * x + d;
}

// determine roots of cubic equation
//   a*x^3 + b*x^2 + c*x + d =0
// algorithm from http://www.1728.com/cubic2.htm
void cubic(double a, double b, double c, double d)
{
    double x1, x2, x3, x1r, x1i=0, x2i, x3i=0;
    cout << a << " " << b << " " << c << " d " << d << endl;
    double f =((3*c/a) - (b*b/(a*a)))/3;
    cout << "f " << f << endl;
    double g=(2*b*b*b/(a*a*a)-9*(b*c/(a*a))+27*d/a)/27;
    cout << "g " << g << endl;
    double h=(g*g)/4+f*f*f/27;
    cout << "h " << h << endl;
    if(h <= 0)
       {
       if(f == 0 && g == 0 &&  h == 0) 
           // all three roots real and equal
           x1 = x2 = x3 =-pow(d/a, 1/3.0);
       else
           // three real roots
           {   
           double i=sqrt((g*g/4)-h);
           cout << "i " << i << endl;
           double j=pow(i, 1/3.0);
           cout << "j " << j << endl;
           double k=acos(-(g/(2*i)));
           cout << "k " << k << endl;
           double l=-j;
           cout << "l " << l << endl;
           double m=cos(k/3);
           cout << "m " << m << endl;
           double n=sqrt(3.0)*sin(k/3);
           cout << "n " << n << endl;
           double p=-(b/(3*a));
           cout << "p " << p << endl;
           x1=2*j*cos(k/3)-(b/(3*a));
           x2=l*(m+n)+p;
           x3=l*(m-n)+p;
           }
       cout << "x1 " << x1 << endl;
       cout << "x2 " << x2 << endl;
       cout << "x3 " <<x3 << endl;
       }
    else
       // one real, two complex roots
       {
        double r=-(g/2)+sqrt(h);
        cout << "r " << r << endl;
        double s=pow(r, 1/3.0);
        cout << "s " << s << endl;
        double t=-(g/2)-sqrt(h);
        cout << "t " << t << endl;
        double u;
        if(t<0) u=-pow(-t, 1/3.0);
        else    u=-pow(t, 1/3.0);
        cout << "u " << u << endl;
        x1=(s+u)-(b/(3*a));
        cout << "x1 " << x1 << endl;
        x2=-(s+u)/2 -(b/(3*a));
        x2i= (s-u)*sqrt(3.0)/2;
        cout << "x2 " << x2 << " +i " << x2i << endl;
        x3=x2;;
        x3i=-x2i;
        cout << "x3 " << x3 << " +i " << x3i << endl;
       }
    cout << "check x1 " << check(a, b, c, d, complex<double>(x1, 0)) << endl;
    cout << "check x2 " << check(a, b, c, d, complex<double>(x2, x3i)) << endl;
    cout << "check x3 " << check(a, b, c, d, complex<double>(x3, x3i)) << endl;
}


int main()
{
    cout << "enter a, b, c, and d ";
    double a, b, c, d;
    cin >> a >> b >> c >> d;
    cubic(a, b, c, d);
    system("pause");
}

try a=3, b=-10, c=14, d= 27 which has one real root and two complex roots -1, 2.166+i2.0749, 2.166-i2.0749