cosh

Devuelve el coseno hiperbólico de un número complejo.

template<class Type> 
   complex<Type> cosh( 
      const complex<Type>& _ComplexNum 
   );

Parámetros

• _ComplexNum
  El número complejo cuyo se está determinando coseno hiperbólico.

Valor devuelto

El número complejo que es el coseno hiperbólico del número complejo de entrada.

Comentarios

Identidades que definen los cosenos hiperbólicos complejos:

cos (z) = (1/2) * (exp (z) + exp (-z))

cos (z) = garrote (a + BI) = garrote (a) cos (b) + sin de isinh (a) (b)

Ejemplo

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

int main( )
{
   using namespace std;
   double pi = 3.14159265359;
   complex <double> c1 ( 3.0 , 4.0 );
   cout << "Complex number c1 = " << c1 << endl;

   // Values of cosine of a complex number c1
   complex <double> c2 = cosh ( c1 );
   cout << "Complex number c2 = cosh ( c1 ) = " << c2 << endl;
   double absc2 = abs ( c2 );
   double argc2 = arg ( c2 );
   cout << "The modulus of c2 is: " << absc2 << endl;
   cout << "The argument of c2 is: "<< argc2 << " radians, which is " 
        << argc2 * 180 / pi << " degrees." << endl << endl; 

   // Hyperbolic cosines of the standard angles 
   // in the first two quadrants of the complex plane
   vector <complex <double> > v1;
   vector <complex <double> >::iterator Iter1;
   complex <double> vc1  ( polar (1.0, pi / 6) );
   v1.push_back( cosh ( vc1 ) );
   complex <double> vc2  ( polar (1.0, pi / 3) );
   v1.push_back( cosh ( vc2 ) );
   complex <double> vc3  ( polar (1.0, pi / 2) );
   v1.push_back( cosh ( vc3) );
   complex <double> vc4  ( polar (1.0, 2 * pi / 3) );
   v1.push_back( cosh ( vc4 ) );
   complex <double> vc5  ( polar (1.0, 5 * pi / 6) );
   v1.push_back( cosh ( vc5 ) );
   complex <double> vc6  ( polar (1.0,  pi ) );
   v1.push_back( cosh ( vc6 ) );

   cout << "The complex components cosh (vci), where abs (vci) = 1"
        << "\n& arg (vci) = i * pi / 6 of the vector v1 are:\n" ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << endl;
}

Requisitos

complejo <deEncabezado: >

Espacio de nombres: std