请帮我把这个程序改成C#的

gaodiancake

//求解任意一元三次方程 a*x^3+b*x^2+c*x+d=0
//输入a,b,c,d;  输出三个解分别为x1,x2,x3
//
#include  <complex>
#include  <iostream>
#include  <math.h>
using   namespace   std;
void  quadratic_equation(double a4,double b4,double c4,complex<double> &z1,complex<double> &z2);
int main()
{
float a,b,c,d;
double p,q,delta;
double M,N;
complex<double> temp1,temp2;
complex<double> x1,x2,x3;
complex<double> y1,y2;

cout<<" please input the coefficients a, b, c, d : ";      
cin>>a>>b>>c>>d;
//输入一元三次方的系数，得方程为a*x^3+b*x^2+c*x+d=0；
   

if(a==0)
{
  quadratic_equation(b,c,d,y1,y2);
  x1 = y1;
  x2 = y2;
  x3 = 0;
  
  std::cout<<" x1="<< x1 <<endl;
  std::cout<<" x2="<< x2 <<endl;
  std::cout<<" x3="<< x3 <<endl;
}
else
{
  
p = -1.0/3*pow((b*1.0/a),2.0)+c*1.0/a;                  
    q = 2.0/27*pow((b*1.0/a),3.0)-1.0/3*b*c/(a*a)+d*1.0/a;  

// p = -1/3*(b/a)^2+c/a;
// q = 2/27*(b/a)^3-1/3*b*c/(a*a)+d/a;
//化成 y^3+p*y+q=0 形式

delta = pow((q/2.0),2.0)+pow((p/3.0),3.0);

    //判别式 delta = (q/2)^2+(p/3)^3;


cout<<" delta>0,有一个实根和两个复根；delta=0,有三个实根；delta<0,有三个不等的实根"<<endl;
    cout<<" 判别式的值 delta="<<delta<<endl;
//delta>0,有一个实根和两个复根；
//delta=0,有三个实根；
    //delta<0,有三个不等的实根。


    complex<double>  omega1(-1.0/2, sqrt(3.0)/2.0);
complex<double>  omega2(-1.0/2, -sqrt(3.0)/2.0);
complex<double>  yy(b/(3.0*a),0.0);
   
    M = -q/2.0;
if(delta<0)
{
   N = sqrt(fabs(delta));
         complex<double>  s1(M,N);  
         complex<double>  s2(M,-N);
   
   x1 = (pow(s1,(1.0/3))+pow(s2,(1.0/3)))-yy;
   x2 = (pow(s1,(1.0/3))*omega1+pow(s2,(1.0/3))*omega2)-yy;
   x3 = (pow(s1,(1.0/3))*omega2+pow(s2,(1.0/3))*omega1)-yy;   
   
  std::cout<<" x1="<< x1 <<endl;
  std::cout<<" x2="<< x2 <<endl;
  std::cout<<" x3="<< x3 <<endl;
  //输出结果 x1=y1-b/(3*a); x2=y2-b/(3*a); x3=y3-b/(3*a);
}
else
{
  N = sqrt(delta);
  //cout<<" please output the M+N="<<M+N<<endl;
  //cout<<" please output the M-N="<<M-N<<endl;
  //  M+N= -q/2+sqrt((q/2)^2+(p/3)^3);
  //  M-N= -q/2-sqrt((q/2)^2+(p/3)^3);
  complex<double>  f1(M+N,0);
  complex<double>  f2(M-N,0);
   if(M+N >= 0)
       temp1 = pow((f1),1.0/3);
   else
    temp1 = -norm(pow(sqrt(f1),1.0/3));

   if(M-N >= 0)
    temp2 = pow((f2),1.0/3);
   else  
    temp2 = -norm(pow(sqrt(f2),1.0/3));

   x1 = temp1+temp2-yy;   
         x2 = omega1*temp1+omega2*temp2-yy;
         x3 = omega2*temp1+omega1*temp2-yy;

   std::cout<<" x1="<< x1 <<endl;
      std::cout<<" x2="<< x2 <<endl;
   std::cout<<" x3="<< x3 <<endl;
  //输出结果 x1=y1-b/(3*a); x2=y2-b/(3*a); x3=y3-b/(3*a);
}
}
  return 0;
}


void  quadratic_equation(double a4,double b4,double c4,complex<double> &z1,complex<double> &z2)
{
  
double delta;
complex<double> temp1,temp2;
delta=b4*b4-4*a4*c4;

complex<double> temp(delta,0);
    temp1 = (-b4)/(2*a4);
temp2 = sqrt(temp)/(2*a4);
z1=temp1+temp2;
z2=temp1-temp2;

}