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Finding the Inverse Hyperbolic Tangent of Complex Number in Golang
  • Last Updated : 27 Mar, 2020

Go language provides inbuilt support for basic constants and mathematical functions for complex numbers with the help of the cmplx package. You are allowed to find the inverse hyperbolic tangent of the specified complex number with the help of the Atanh() function provided by the math/cmplx package. So, you need to add a math/cmplx package in your program with the help of the import keyword to access the Atanh() function.

Syntax:

func Atanh(x complex128) complex128

Let us discuss this concept with the help of the given examples:

Example 1:

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// Golang program to illustrate how to find the
// Inverse Hyperbolic Tangent of Complex Number
  
package main
  
import (
    "fmt"
    "math/cmplx"
)
  
// Main function
func main() {
  
    // Finding the inverse hyperbolic tangent 
    // of the specified complex number
    // Using Atanh() function
    res_1 := cmplx.Atanh(2 + 5i)
    res_2 := cmplx.Atanh(-9 + 8i)
    res_3 := cmplx.Atanh(-5 - 7i)
  
    // Displaying the result
    fmt.Println("Result 1:", res_1)
    fmt.Println("Result 2:", res_2)
    fmt.Println("Result 3:", res_3)
}

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Output:

Result 1: (0.06706599664866984+1.3992843565845448i)
Result 2: (-0.0619590409761453+1.5154677162079488i)
Result 3: (-0.06706599664866986-1.4760562478543229i)

Example 2 :

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// Golang program to illustrate how to find
// Inverse Hyperbolic Tangent of Complex Number
package main
  
import (
    "fmt"
    "math/cmplx"
)
  
// Main function
func main() {
  
    cnumber_1 := complex(5, 7)
    cnumber_2 := complex(6, 9)
  
    // Finding inverse hyperbolic tangent
    cvalue_1 := cmplx.Atanh(cnumber_1)
    cvalue_2 := cmplx.Atanh(cnumber_2)
  
    // Sum of two inverse hyperbolic tangent values
    res := cvalue_1 + cvalue_2
  
    // Displaying results
    fmt.Println("Complex Number 1: ", cnumber_1)
    fmt.Println("Inverse hyperbolic tangent 1: ", cvalue_1)
  
    fmt.Println("Complex Number 2: ", cnumber_2)
    fmt.Println("Inverse hyperbolic tangent 2: ", cvalue_2)
    fmt.Println("Sum of inverse hyperbolic tangents : ", res)
  
}

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Output:

Complex Number 1:  (5+7i)
Inverse hyperbolic tangent 1:  (0.06706599664866984+1.4760562478543229i)
Complex Number 2:  (6+9i)
Inverse hyperbolic tangent 2:  (0.051023839085878805+1.4938239945657217i)
Sum of inverse hyperbolic tangents :  (0.11808983573454865+2.969880242420045i)



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