cosh

Article
06/09/2015

Renvoie le cosinus hyperbolique d'un nombre complexe.

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

Paramètres

_ComplexNum
Le nombre complexe dont le cosinus hyperbolique est déterminé.

Valeur de retour

Le nombre complexe qui est le cosinus hyperbolique du nombre complexe d'entrée.

Notes

Définition des identités cosinus hyperboliques complexes :

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

cos (z) = cosh (a + bi) = cosh (a) cos (b) + isinh (a) sin (b)

Exemple

// 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;
}

Configuration requise

**En-tête :**complexe <de >

Espace de noms : std

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