[Bug math/17118] New: ctanh(INFINITY + 2 * I) returns incorrect value

["kariya_mitsuru at hotmail dot com" <sourceware-bugzilla at sourceware dot org>]

https://sourceware.org/bugzilla/show_bug.cgi?id=17118

            Bug ID: 17118
           Summary: ctanh(INFINITY + 2 * I) returns incorrect value
           Product: glibc
           Version: unspecified
            Status: NEW
          Severity: normal
          Priority: P2
         Component: math
          Assignee: unassigned at sourceware dot org
          Reporter: kariya_mitsuru at hotmail dot com

ctanh(INFINITY + 2 * I) returns 1 + 0 * I, but it should return 1 - 0 * I.

According to C99(or C11) standard, ctanh(INFINITY + y * I) should return 1.0 +
I * +0.0 * sin(2.0 * y) for positive-signed finite y if an implementation
defines __STDC_IEC_559_COMPLEX__.
If y is 2, sin(2.0 * y) returns about -0.756802 so "+0.0 * sin(2.0 * y)" is
-0.0.

==== test source ========================================
#include <stdio.h>
#include <complex.h>
#include <math.h>

int main()
{
    for (int i = 0; i < 5; i++) {
        const complex double c = +INFINITY + i * I;
        const complex double r1 = tanh(c);
        const double r2 = sin(2.0 * cimag(c));
        printf("ctanh(%g + %gi) = %g + %gi, sin(2 * %g) = %g\n", creal(c),
cimag(c), creal(r1), cimag(r1), cimag(c), r2);
    }
}
==== output =============================================
ctanh(inf + 0i) = 1 + 0i, sin(2 * 0) = 0
ctanh(inf + 1i) = 1 + 0i, sin(2 * 1) = 0.909297
ctanh(inf + 2i) = 1 + 0i, sin(2 * 2) = -0.756802
ctanh(inf + 3i) = 1 + 0i, sin(2 * 3) = -0.279415
ctanh(inf + 4i) = 1 + 0i, sin(2 * 4) = 0.989358

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