ctanhf, ctanh, ctanhl - cppreference.com (original) (raw)

Defined in header <complex.h>
float complex ctanhf( float complex z ); (1) (since C99)
double complex ctanh( double complex z ); (2) (since C99)
long double complex ctanhl( long double complex z ); (3) (since C99)
Defined in header <tgmath.h>
#define tanh( z ) (4) (since C99)

1-3) Computes the complex hyperbolic tangent of z.

  1. Type-generic macro: If z has type long double complex, ctanhl is called. if z has type double complex, ctanh is called, if z has type float complex, ctanhf is called. If z is real or integer, then the macro invokes the corresponding real function (tanhf, tanh, tanhl). If z is imaginary, then the macro invokes the corresponding real version of the function tan, implementing the formula tanh(iy) = i tan(y), and the return type is imaginary.

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[edit] Parameters

[edit] Return value

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

[edit] Error handling and special values

Errors are reported consistent with math_errhandling

If the implementation supports IEEE floating-point arithmetic,

  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

[edit] Notes

Mathematical definition of hyperbolic tangent is tanh z =

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.

[edit] 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)); }

Output:

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

[edit] References

[edit] See also