Finding the Inverse Cosine of Complex Number in Golang
Last Updated :
26 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 cosine of the specified complex number with the help of Acos() 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 Acos() function.
Syntax:
func Acos(x complex128) complex128
Let us discuss this concept with the help of the given examples:
Example 1 :
package main
import (
"fmt"
"math/cmplx"
)
func main() {
res_1 := cmplx.Acos(3 + 5i)
res_2 := cmplx.Acos(-4 + 8i)
res_3 := cmplx.Acos(-8 - 7i)
fmt.Println( "Result 1:" , res_1)
fmt.Println( "Result 2:" , res_2)
fmt.Println( "Result 3:" , res_3)
}
|
Output:
Result 1: (1.0367972572007265-2.4598315216234306i)
Result 2: (2.031957962047241-2.886039504947561i)
Result 3: (2.4205679446613617+3.0565545070216835i)
Example 2 :
package main
import (
"fmt"
"math/cmplx"
)
func main() {
cnumber_1 := complex(5, 7)
cnumber_2 := complex(6, 9)
cvalue_1 := cmplx.Acos(cnumber_1)
cvalue_2 := cmplx.Acos(cnumber_2)
res := cvalue_1 + cvalue_2
fmt.Println( "Complex Number 1: " , cnumber_1)
fmt.Println( "Inverse Cosine 1: " , cvalue_1)
fmt.Println( "Complex Number 2: " , cnumber_2)
fmt.Println( "Inverse Cosine 2: " , cvalue_2)
fmt.Println( "Final sum: " , res)
}
|
Output:
Complex Number 1: (5+7i)
Inverse Cosine 1: (0.9537320301189085-2.846288828208389i)
Complex Number 2: (6+9i)
Inverse Cosine 2: (0.9847612348751953-3.075060767946888i)
Final sum: (1.9384932649941038-5.921349596155277i)
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