polar

返回，它对应的指定的模数和参数，以解析形式。

template<class Type> 
   complex<Type> polar( 
      const Type& _Modulus, 
      const Type& _Argument = 0 
   );

参数

• _Modulus
  复数的模数输入。

• _Argument
  复数的参数输入。

返回值

在极的复数形式指定的解决窗体。

备注

复数的极窗体的模数 r 和参数，这些参数与实际和虚部的解决组件 a 和 b 由公式 的 = r *，因此 (=) 和 b r 给定 (*)。

示例

// complex_polar.cpp
// compile with: /EHsc
#include <complex>
#include <iostream>

int main( )
{
   using namespace std;
   double pi = 3.14159265359;

   // Complex numbers can be entered in polar form with
   // modulus and argument parameter inputs but are
   // stored in Cartesian form as real & imag coordinates
   complex <double> c1 ( polar ( 5.0 ) );   // Default argument = 0
   complex <double> c2 ( polar ( 5.0 , pi / 6 ) );
   complex <double> c3 ( polar ( 5.0 , 13 * pi / 6 ) );
   cout << "c1 = polar ( 5.0 ) = " << c1 << endl;
   cout << "c2 = polar ( 5.0 , pi / 6 ) = " << c2 << endl;
   cout << "c3 = polar ( 5.0 , 13 * pi / 6 ) = " << c3 << endl;

   if ( (arg ( c2 ) <= ( arg ( c3 ) + .00000001) ) || 
        (arg ( c2 ) >= ( arg ( c3 ) - .00000001) ) )
      cout << "The complex numbers c2 & c3 have the "
           << "same principal arguments."<< endl;
   else
      cout << "The complex numbers c2 & c3 don't have the "
           << "same principal arguments." << endl;

   // the modulus and argument of a complex number can be rcovered
   double absc2 = abs ( c2 );
   double argc2 = arg ( c2 );
   cout << "The modulus of c2 is recovered from c2 using: abs ( c2 ) = "
        << absc2 << endl;
   cout << "Argument of c2 is recovered from c2 using:\n arg ( c2 ) = "
        << argc2 << " radians, which is " << argc2 * 180 / pi
        << " degrees." << endl; 
}

要求

复杂页眉： <>

命名空间: std