ctanhf, ctanh, ctanhl

Parameters

Return value

If no errors occur, complex hyperbolic tangent of z is returned

Error handling and special values

Errors are reported consistent with math_errhandling

If the implementation supports IEEE floating-point arithmetic,

• ctanh ( conj ( z ) ) == conj ( ctanh ( z ) )
• ctanh ( - z ) == - ctanh ( z )
• If z is +0+0i , the result is +0+0i
• If z is x+∞i (for any ^[1] finite x), the result is NaN+NaNi and FE_INVALID is raised
• If z is x+NaN (for any ^[2] finite x), the result is NaN+NaNi and FE_INVALID may be raised
• If z is +∞+yi (for any finite positive y), the result is 1+0i
• If z is +∞+∞i , the result is 1±0i (the sign of the imaginary part is unspecified)
• If z is +∞+NaNi , the result is 1±0i (the sign of the imaginary part is unspecified)
• If z is NaN+0i , the result is NaN+0i
• If z is NaN+yi (for any non-zero y), the result is NaN+NaNi and FE_INVALID may be raised
• If z is NaN+NaNi , the result is NaN+NaNi

1. ↑ per DR471 , this only holds for non-zero x. If z is 0+∞i , the result should be 0+NaNi
2. ↑ per DR471 , this only holds for non-zero x. If z is 0+NaNi , the result should be 0+NaNi

Notes

Hyperbolic tangent is an analytical function on the complex plane and has no branch cuts. It is periodic with respect to the imaginary component, with period πi, and has poles of the first order along the imaginary line, at coordinates (0, π(1/2 + n)) . However no common floating-point representation is able to represent π/2 exactly, thus there is no value of the argument for which a pole error occurs.

Example

#include <stdio.h>
#include <math.h>
#include <complex.h>
 
int main(void)
{
    double complex z = ctanh(1);  // behaves like real tanh along the real line
    printf("tanh(1+0i) = %f%+fi (tanh(1)=%f)\n", creal(z), cimag(z), tanh(1));
 
    double complex z2 = ctanh(I); // behaves like tangent along the imaginary line
    printf("tanh(0+1i) = %f%+fi ( tan(1)=%f)\n", creal(z2), cimag(z2), tan(1));
}

tanh(1+0i) = 0.761594+0.000000i (tanh(1)=0.761594)
tanh(0+1i) = 0.000000+1.557408i ( tan(1)=1.557408)

References