tanh

傳回複數的雙曲線正切函數。

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

參數

• _ComplexNum
  的雙曲線正切函數來決定的複數。

傳回值

是輸入複數的雙曲線正切函數的複數。

備註

定義複雜 Hyperbolic Cotangent 的識別:

tanh (z) = sinh (z)/短棒 (z) = (exp (z) – exp (-z)/(exp (z) + exp (-z))

範例

// complex_tanh.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 = tanh ( c1 );
   cout << "Complex number c2 = tanh ( 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 tangents 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( tanh ( vc1 ) );
   complex <double> vc2  ( polar ( 1.0, pi / 3 ) );
   v1.push_back( tanh ( vc2 ) );
   complex <double> vc3  ( polar ( 1.0, pi / 2 ) );
   v1.push_back( tanh ( vc3 ) );
   complex <double> vc4  ( polar ( 1.0, 2 * pi / 3 ) );
   v1.push_back( tanh ( vc4 ) );
   complex <double> vc5  ( polar ( 1.0, 5 * pi / 6 ) );
   v1.push_back( tanh ( vc5 ) );
   complex <double> vc6  ( polar ( 1.0, pi ) );
   v1.push_back( tanh ( vc6 ) );

   cout << "The complex components tanh (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;
}

需求

標題: <複雜>

命名空間: std