std::pow(std::complex) (3) - Linux Manuals

std::pow(std::complex): std::pow(std::complex)

NAME

std::pow(std::complex) - std::pow(std::complex)

Synopsis

Defined in header <complex>
template< class T > (1)
complex<T> pow( const complex<T>& x, const complex<T>& y);
template< class T > (2)
complex<T> pow( const complex<T>& x, const T& y);
template< class T > (3)
complex<T> pow( const T& x, const complex<T>& y);
template< class T, class U > (4) (since C++11)
complex</*Promoted*/> pow( const complex<T>& x, const complex<U>& y);
template< class T, class U > (5) (since C++11)
complex</*Promoted*/> pow( const complex<T>& x, const U& y);
template< class T, class U > (6) (since C++11)
complex</*Promoted*/> pow( const T& x, const complex<U>& y);

1-3) Computes complex x raised to a complex power y with a branch cut along the negative real axis for the first argument.

4-6) Additional overloads are provided for all arithmetic types, such that

      1. If either argument is long double or std::complex<long double>, then both arguments are cast to std::complex<long double> (since C++11)
      2. Otherwise, if either argument is double, std::complex<double> or integer type, then both arguments are cast to std::complex<double>
      3. Otherwise, if either argument is float or std::complex<float>, then both arguments are cast to std::complex<float>

Parameters

x - base as a complex value
y - exponent as a complex value

Return value

If no errors occur, the complex power xy
, is returned.
Errors and special cases are handled as if the operation is implemented by std::exp(y*std::log(x))
The result of std::pow(0, 0) is implementation-defined.

Example

// Run this code

  #include <iostream>
  #include <complex>

  int main()
  {
      std::cout << std::fixed;

      std::complex<double> z(1, 2);
      std::cout << "(1,2)^2 = " << std::pow(z, 2) << '\n';

      std::complex<double> z2(-1, 0); // square root of -1
      std::cout << "-1^0.5 = " << std::pow(z2, 0.5) << '\n';

      std::complex<double> z3(-1, -0.0); // other side of the cut
      std::cout << "(-1, -0)^0.5 = " << std::pow(z3, 0.5) << '\n';

      std::complex<double> i(0, 1); // i^i = exp(-pi/2)
      std::cout << "i^i = " << std::pow(i, i) << '\n';
  }

Output:

  (1,2)^2 = (-3.000000,4.000000)
  -1^0.5 = (0.000000,1.000000)
  (-1, -0)^0.5 = (0.000000,-1.000000)
  i^i = (0.207880,0.000000)