std::atanh(std::complex)

Defined in header `<complex>`
template< class T > complex<T> atanh( const complex<T>& z );		(since C++11)

Computes the complex arc hyperbolic tangent of z with branch cuts outside the interval $[1; +1]$ along the real axis.

Parameters

z

-

complex value

Return value

If no errors occur, the complex arc hyperbolic tangent of z is returned, in the range of a half-strip mathematically unbounded along the real axis and in the interval $[i/2; +i/2]$ along the imaginary axis.

Error handling and special values

Errors are reported consistent with math_errhandling

If the implementation supports IEEE floating-point arithmetic,

std::atanh(std::conj(z)) == std::conj(std::atanh(z))
std::atanh(-z) == -std::atanh(z)
If z is (+0,+0), the result is (+0,+0)
If z is (+0,NaN), the result is (+0,NaN)
If z is (+1,+0), the result is (+,+0) and FE_DIVBYZERO is raised
If z is (x,+) (for any finite positive x), the result is (+0,/2)
If z is (x,NaN) (for any finite nonzero x), the result is (NaN,NaN) and FE_INVALID may be raised
If z is (+,y) (for any finite positive y), the result is (+0,/2)
If z is (+,+), the result is (+0,/2)
If z is (+,NaN), the result is (+0,NaN)
If z is (NaN,y) (for any finite y), the result is (NaN,NaN) and FE_INVALID may be raised
If z is (NaN,+), the result is (0,/2) (the sign of the real part is unspecified)
If z is (NaN,NaN), the result is (NaN,NaN)

Notes

Although the C++ standard names this function "complex arc hyperbolic tangent", the inverse functions of the hyperbolic functions are the area functions. Their argument is the area of a hyperbolic sector, not an arc. The correct name is "complex inverse hyperbolic tangent", and, less common, "complex area hyperbolic tangent".

Inverse hyperbolic tangent is a multivalued function and requires a branch cut on the complex plane. The branch cut is conventionally placed at the line segmentd $(-,-1]$ and $[+1,+)$ of the real axis.

The mathematical definition of the principal value of the inverse hyperbolic tangent is

atanh z = ln(1+z)-ln(1-z) 2

.

For any z,

atanh(z) = atan(iz) i

Example

Run this code

#include <iostream>
#include <complex>

int main()
{
    std::cout << std::fixed;
    std::complex<double> z1(2, 0);
    std::cout << "atanh" << z1 << " = " << std::atanh(z1) << '\n';

    std::complex<double> z2(2, -0.0);
    std::cout << "atanh" << z2 << " (the other side of the cut) = "
              << std::atanh(z2) << '\n';

    // for any z, atanh(z) = atanh(iz)/i
    std::complex<double> z3(1,2);
    std::complex<double> i(0,1);
    std::cout << "atanh" << z3 << " = " << std::atanh(z3) << '\n'
              << "atan" << z3*i << "/i = " << std::atan(z3*i)/i << '\n';
}

Output:

atanh(2.000000,0.000000) = (0.549306,1.570796)
atanh(2.000000,-0.000000) (the other side of the cut) = (0.549306,-1.570796)
atanh(1.000000,2.000000) = (0.173287,1.178097)
atan(-2.000000,1.000000)/i = (0.173287,1.178097)

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