Leave sign of 0 unspecified for tanh `inf + yj`

[ndgrigorian]

The array API spec states, for tanh complex special cases

If a is +infinity and b is a positive (i.e., greater than 0) finite number, the result is 1 + 0j

from open-std pages describing ctanh

— ctanh(+∞ + iy) returns 1 + i0sin(2y), for positive-signed finite y.

This matches up to the math (we can derive this from tanh(x + iy) taking the limit as x approaches infinity and constraining y to be finite), and gives us 1 ± 0i, with the sign being inherited from sin(2y) if we care about signed zeros, rather than positive zero in all cases.

This was brought up because this special case is tested by the array API tests, which causes dpnp to fail the test.

[kgryte]

Thanks for raising this issue, @ndgrigorian. For complex number special handling, we deferred to behavior described in C99: https://en.cppreference.com/c/numeric/complex/ctanh. Namely,

If z is +∞+yi (for any finite positive y), the result is 1+0i

[ndgrigorian]

Thanks for raising this issue, @ndgrigorian. For complex number special handling, we deferred to behavior described in C99: https://en.cppreference.com/c/numeric/complex/ctanh. Namely,

If z is +∞+yi (for any finite positive y), the result is 1+0i

yes, but I would argue, that the cppreference page is a bit misleading as well. The ISO standard (final draft) text for C99 also uses sin(2y).

NumPy also gives the same:

import numpy as np

print([np.tanh(complex(np.inf, n)) for n in range(4)])
# [np.complex128(1+0j), np.complex128(1+0j), np.complex128(1-0j), np.complex128(1-0j)]

at least for conda-forge NumPy, there can be discrepancies on PyPI with complex edge cases I've found.