在C ++中具有复数的方程给出错误的结果

Question

Hi I'm getting the wrong expected result and I think it's do to std::complex, heres the result I should be getting according to matlab and here's the result if you run the code below this is the result i get , basically everything is just NAN, where did i go wrong?

#include<cmath>
#include<complex>
#include<new>
#include<iostream>



    int sign(double x){
        if(x > 0)    {
            return 1;
        }
        else if(x < 0)    {
            return -1;
        }
        else    {
            return 0;
        }
    }

    int main(){

      double alpha = 1.8;
      double beta = .35;
      double sigma = 1;
      double mu = 0.5;
      double PI = 3.1416; 
      int N = 8192;
      double h = 0.01;
      std::complex<double>phi[N]; 
      double* in_t2= new double[N]; 


     double* abs_t = new double [N]; 
     double* sign_t = new double [N];

     std::complex<double> I(0,1);
     double s = 0.01;

     s = 1/(h*N);
     std::cout<<s;
     for (int i=1; i<=N; i++) {
       in_t2[i-1] = 2*PI * (i - 1 - N/2)*s; // x1

     }

     for (int i = 0; i < N; i++){
       abs_t[i] = std::abs(in_t2[i]);
     }

     for (int i = 0; i < N; i++){
       sign_t[i] = sign(in_t2[i]);
     }

      for (int i = 0; i < N; i++){ 
        //where i suspect the error is..


        phi[i] = pow(abs_t[i],sign_t[i]);
       }


      for ( int i = 0; i< N; i++){
        if (in_t2[i] == 0){
          phi[i] = 0;
        }
        phi[i] = std::exp(phi[i]);
      }


      return 0;
    }

Answer 1

The pow function called here

pow((-1.0*abs_t[i]), alpha)

is the one for double arguments and results, which returns a NaN for negative bases.

Answer 2

std::pow will return the same type that is passed to it.

If you need a std::complex to be returned from std::pow() , you must pass a std::complex to the function.

You can easily solve this by casting any expression

pow( expression )

to:

pow( std::complex( expression ) )