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Roots of Unity

Given a small integer n, print all the n’th roots of unity up to 6 significant digits. We basically need to find all roots of equation xn – 1.

Examples:

Input :  n = 1
Output : 1.000000 + i 0.000000
x - 1 = 0 , has only one root i.e., 1

Input :  2
Output : 1.000000 + i 0.000000
    -1.000000 + i 0.000000
x2 - 1 = 0 has 2 distinct roots, i.e., 1 and -1 


Any complex number is said to be root of unity if it gives 1 when raised to some power.

nth root of unity is any complex number such that it gives 1 when raised to the power n.

Mathematically, 
An nth root of unity, where n is a positive integer 
(i.e. n = 1, 2, 3, …) is a number z satisfying the
equation 

z^n  = 1
or , 
z^n - 1 = 0

We can use the De Moivre’s formula here ,

( Cos x + i Sin x )^k = Cos kx + i Sin kx

Setting x = 2*pi/n, we can obtain all the nth roots 
of unity, using the fact that Nth roots are set of 
numbers given by,

Cos (2*pi*k/n) + i Sin(2*pi*k/n)
Where, 0 <= k < n

Using the above fact we can easily print all the nth roots of unity !
Below is the program for the same.

C++

// C++ program to print n'th roots of unity
#include <bits/stdc++.h>
using namespace std;
  
// This function receives an integer n , and prints
// all the nth roots of unity
void printRoots(int n)
{
    // theta = 2*pi/n
    double theta = M_PI*2/n;
  
    // print all nth roots with 6 significant digits
    for(int k=0; k<n; k++)
    {
        // calculate the real and imaginary part of root
        double real = cos(k*theta);
        double img = sin(k*theta);
  
        // Print real and imaginary parts
        printf("%.6f", real);
        img >= 0? printf(" + i "): printf(" - i ");
        printf("%.6f ", abs(img));
    }
}
  
// Driver function to check the program
int main()
{
    printRoots(1);
    cout << endl;
    printRoots(2);
    cout << endl;
    printRoots(3);
    return 0;
}

Java

// Java program to print n'th roots of unity
import java.io.*;
  
class GFG {
  
// This function receives an integer n , and prints
// all the nth roots of unity
static void printRoots(int n)
{
    // theta = 2*pi/n
    double theta = 3.14*2/n;
  
    // print all nth roots with 6 significant digits
    for(int k=0; k<n; k++)
    {
        // calculate the real and imaginary part of root
        double real = Math.cos(k*theta);
        double img = Math.sin(k*theta);
  
        // Print real and imaginary parts
        System.out.println(real);
        if (img >= 0)
            System.out.println(" + i ");
        else
            System.out.println(" - i ");
        System.out.println(Math.abs(img));
    }
}
  
// Driver function to check the program
public static void main (String[] args)
{
    printRoots(1);
    //System.out.println();
    printRoots(2);
    //System.out.println();
    printRoots(3);
}
}
// This code is conributed by Raj

Python3

# Python3 program to print n'th roots of unity
  
import math
  
# This function receives an integer n , and prints
# all the nth roots of unity
def printRoots(n):
  
    # theta = 2*pi/n
    theta = math.pi * 2 / n
  
    # print all nth roots with 6 significant digits
    for k in range(0, n):
  
        # calculate the real and imaginary part of root
        real = math.cos(k * theta)
        img = math.sin(k * theta)
  
        # Print real and imaginary parts
        print(real, end=" ")
        if(img >= 0):
            print(" + i ", end=" ")
        else:
            print(" - i ", end=" ")
        print(abs(img))
  
  
# Driver function to check the program
if __name__=='__main__':
    printRoots(1)
    printRoots(2)
    printRoots(3)
  
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to print n'th roots of unity 
using System;
  
class GFG { 
  
// This function receives an integer n , and prints 
// all the nth roots of unity 
static void printRoots(int n) 
    // theta = 2*pi/n 
    double theta = 3.14*2/n; 
  
    // print all nth roots with 6 significant digits 
    for(int k=0; k<n; k++) 
    
        // calculate the real and imaginary part of root 
        double real = Math.Cos(k*theta); 
        double img = Math.Sin(k*theta); 
  
        // Print real and imaginary parts 
        Console.Write(real); 
        if (img >= 0) 
            Console.Write(" + i "); 
        else
            Console.Write(" - i "); 
        Console.WriteLine(Math.Abs(img)); 
    
  
// Driver function to check the program 
static void Main() 
    printRoots(1); 
       
    printRoots(2); 
       
    printRoots(3); 
// This code is conributed by mits

PHP

= 0 ? print(” + i “): print(” – i “);
printf(round(abs($img), 6) . “ ”);
}
}

// Driver Code
printRoots(1);
printRoots(2);
printRoots(3);

// This code is contributed by mits
?>


Output:

1.000000 + i 0.000000
1.000000 + i 0.000000
-1.000000 + i 0.000000
1.000000 + i 0.000000
-0.500000 + i 0.866025
-0.500000 - i 0.866025

References : Wikipedia

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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