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在头文件<math.h>中定义



float complex       casinhf( float complex z );

(1)

(since C99)

double complex      casinh( double complex z );

(2)

(since C99)

long double complex casinhl( long double complex z );

(3)

(since C99)

Defined in header <tgmath.h>



#define asinh( z )

(4)

(since C99)

1-3）计算z间隔-i以外的分支切割的复数圆弧双曲正弦; + i沿着虚轴。

4）类型 - 通用宏：如果z有类型long double complex，casinhl被调用。如果z有类型double complex，casinh称为，如果z有类型float complex，casinhf称为。如果z是真实的或整数，则宏调用相应的实函数（asinhf，asinh，asinhl）。如果z是虚构的，那么宏调用函数的相应实际版本asin，实现公式asinh（iy）= i asin（y），并且返回类型是虚数。

参数

z

-

复杂的论点

返回值

如果没有出现错误，z则返回复数双曲正弦曲线，位于实数轴无数学条带和-iπ/ 2区间范围内; +iπ/ 2沿着虚轴。

错误处理和特殊值

报告的错误与math_errhandling一致。

如果实现支持IEEE浮点运算，

  • casinh(conj(z)) == conj(casinh(z))

  • casinh(-z) == -casinh(z)

  • 如果z是+0+0i，结果是+0+0i

  • 如果z是x+∞i（对于任何正的有限x），结果是+∞+π/2

  • 如果z是x+NaNi（对于任何有限的x），结果是NaN+NaNi并FE_INVALID可能会提出

  • 如果z是+∞+yi（对于任何有限的y），结果是+∞+0i

  • 如果z是+∞+∞i，结果是+∞+iπ/4

  • 如果z是+∞+NaNi，结果是+∞+NaNi

  • 如果z是NaN+0i，结果是NaN+0i

  • 如果z是NaN+yi（对于任何有限的非零y），结果是NaN+NaNi并FE_INVALID可能会提出

  • 如果z是NaN+∞i，结果是±∞+NaNi（真实部分的符号未指定）

  • 如果z是NaN+NaNi，结果是NaN+NaNi

笔记

尽管C标准将这个函数命名为“复曲线双曲正弦”，但双曲函数的反函数是区域函数。他们的论点是双曲线领域，而不是弧线。正确的名称是“复数反双曲正弦”，不太常见的是“复曲面双曲正弦”。

反双曲正弦是一种多值函数，需要在复平面上进行分支切分。分支切割通常放置在线段（ - 我 - ∞，我）和（我，我虚轴的∞）。

反双曲正弦的主值的数学定义为asinh z = ln（z +√1+ z2

）对于asinh（z）=中的任何一个

| asin(iz) |

|:----|

| i |

例

#include <stdio.h>#include <complex.h>
 int main(void){
    double complex z = casinh(0+2*I);    printf("casinh(+0+2i) = %f%+fi\n", creal(z), cimag(z));
 
    double complex z2 = casinh(-conj(2*I)); // or casinh(CMPLX(-0.0, 2)) in C11    printf("casinh(-0+2i) (the other side of the cut) = %f%+fi\n", creal(z2), cimag(z2)); 
    // for any z, asinh(z) = asin(iz)/i
    double complex z3 = casinh(1+2*I);    printf("casinh(1+2i) = %f%+fi\n", creal(z3), cimag(z3));
    double complex z4 = casin((1+2*I)*I)/I;    printf("casin(i * (1+2i))/i = %f%+fi\n", creal(z4), cimag(z4));}

输出：

casinh(+0+2i) = 1.316958+1.570796icasinh(-0+2i) (the other side of the cut) = -1.316958+1.570796icasinh(1+2i) = 1.469352+1.063440icasin(i * (1+2i))/i =  1.469352+1.063440i

参考

  • C11标准（ISO / IEC 9899：2011）：

    • 7.3.6.2 casinh函数（p：192-193）

    • 7.25类型通用数学<tgmath.h>（p：373-375）

    • G.6.2.2赌场功能（p：540）

    • G.7类型 - 通用数学<tgmath.h>（p：545）

  • C99标准（ISO / IEC 9899：1999）：

    • 7.3.6.2赌场功能（p：174-175）

    • 7.22类型通用数学<tgmath.h>（p：335-337）

    • G.6.2.2赌场功能（p：475）

    • G.7类型 - 通用数学<tgmath.h>（p：480）

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