Square root of two Complex Numbers

Given two positive integers A and B representing the complex number Z in the form of Z = A + i * B, the task is to find the square root of the given complex number.

Examples:

Input: A = 0, B =1
Output:
The Square roots are:
0.707107 + 0.707107*i
-0.707107 - 0.707107*i

Input: A = 4, B = 0
Output:
The Square roots are:
2
-2

Approach: The given problem can be solved based on the following observations:

It is known that the square root of a complex number is also a complex number.
Then considering the square root of the complex number equal to X + i*Y, the value of (A + i*B) can be expressed as:
- A + i * B = (X + i * Y) * (X + i * Y)
- A + i * B = X² - Y²+ 2 * i * X * Y
Equating the value of real and complex parts individually:
- X = \sqrt (\frac {(A \pm \sqrt(A^{2} + B^{2}))}{2})
- Y = \frac{B}{2 \times \sqrt (\frac {(A \pm \sqrt(A^{2} + B^{2}))}{2})}

From the above observations, calculate the value of X and Y using the above formula and print the value (X + i*Y) as the resultant square root value of the given complex number.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the square root of
// a complex number
void complexRoot(int A, int B)
{
    // Stores all the square roots
    vector<pair<double, double> > ans;

    // Stores the first square root
    double X1 = abs(sqrt((A + sqrt(A * A
                                   + B * B))
                         / 2));
    double Y1 = B / (2 * X1);

    // Push the square root in the ans
    ans.push_back({ X1, Y1 });

    // Stores the second square root
    double X2 = -1 * X1;
    double Y2 = B / (2 * X2);

    // If X2 is not 0
    if (X2 != 0) {

        // Push the square root in
        // the array ans[]
        ans.push_back({ X2, Y2 });
    }

    // Stores the third square root
    double X3 = (A - sqrt(A * A + B * B)) / 2;

    // If X3 is greater than 0
    if (X3 > 0) {
        X3 = abs(sqrt(X3));
        double Y3 = B / (2 * X3);

        // Push the square root in
        // the array ans[]
        ans.push_back({ X3, Y3 });

        // Stores the fourth square root
        double X4 = -1 * X3;
        double Y4 = B / (2 * X4);

        if (X4 != 0) {

            // Push the square root
            // in the array ans[]
            ans.push_back({ X4, Y4 });
        }
    }

    // Prints the square roots
    cout << "The Square roots are: "
         << endl;

    for (auto p : ans) {
        cout << p.first;
        if (p.second > 0)
            cout << "+";
        if (p.second)
            cout << p.second
                 << "*i" << endl;
        else
            cout << endl;
    }
}

// Driver Code
int main()
{
    int A = 0, B = 1;
    complexRoot(A, B);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{
    
static class pair
{ 
    double first, second; 
    public pair(double first,
                double second)  
    { 
        this.first = first; 
        this.second = second; 
    }    
} 

// Function to find the square root of
// a complex number
static void complexRoot(int A, int B)
{
    
    // Stores all the square roots
    Vector<pair> ans = new Vector<pair>();

    // Stores the first square root
    double X1 = Math.abs(Math.sqrt((A + Math.sqrt(A * A + 
                                    B * B)) / 2));
    double Y1 = B / (2 * X1);

    // Push the square root in the ans
    ans.add(new pair( X1, Y1 ));

    // Stores the second square root
    double X2 = -1 * X1;
    double Y2 = B / (2 * X2);

    // If X2 is not 0
    if (X2 != 0) 
    {
        
        // Push the square root in
        // the array ans[]
        ans.add(new pair(X2, Y2));
    }

    // Stores the third square root
    double X3 = (A - Math.sqrt(A * A + B * B)) / 2;

    // If X3 is greater than 0
    if (X3 > 0) 
    {
        X3 = Math.abs(Math.sqrt(X3));
        double Y3 = B / (2 * X3);

        // Push the square root in
        // the array ans[]
        ans.add(new pair(X3, Y3));

        // Stores the fourth square root
        double X4 = -1 * X3;
        double Y4 = B / (2 * X4);

        if (X4 != 0)
        {
            
            // Push the square root
            // in the array ans[]
            ans.add(new pair(X4, Y4));
        }
    }

    // Prints the square roots
    System.out.print("The Square roots are: " + "\n");

    for(pair p : ans)
    {
        System.out.printf("%.4f", p.first);
        if (p.second > 0)
            System.out.print("+");
        if (p.second != 0)
            System.out.printf("%.4f*i\n", p.second);
        else
            System.out.println();
    }
}

// Driver Code
public static void main(String[] args)
{
    int A = 0, B = 1;
    
    complexRoot(A, B);
}
}

// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
from math import sqrt

# Function to find the square root of
# a complex number
def complexRoot(A, B):
    
    # Stores all the square roots
    ans = []

    # Stores the first square root
    X1 = abs(sqrt((A + sqrt(A * A + B * B)) / 2))
    Y1 = B / (2 * X1)

    # Push the square root in the ans
    ans.append([X1, Y1])

    # Stores the second square root
    X2 = -1 * X1
    Y2 = B / (2 * X2)

    # If X2 is not 0
    if (X2 != 0):

        # Push the square root in
        # the array ans[]
        ans.append([X2, Y2])

    # Stores the third square root
    X3 = (A - sqrt(A * A + B * B)) / 2

    # If X3 is greater than 0
    if (X3 > 0):
        X3 = abs(sqrt(X3))
        Y3 = B / (2 * X3)

        # Push the square root in
        # the array ans[]
        ans.append([X3, Y3])

        # Stores the fourth square root
        X4 = -1 * X3
        Y4 = B / (2 * X4)

        if (X4 != 0):

            # Push the square root
            # in the array ans[]
            ans.append([X4, Y4])

    # Prints the square roots
    print("The Square roots are: ")

    for p in ans:
        print(round(p[0], 6), end = "")
        if (p[1] > 0):
            print("+", end = "")
        if (p[1]):
            print(str(round(p[1], 6)) + "*i")
        else:
            print()

# Driver Code
if __name__ == '__main__':
    
    A,B = 0, 1
    complexRoot(A, B)

# This code is contributed by mohit kumar 29

// C# code to implement the approach
using System;
using System.Collections.Generic;

class GFG
{
  // Definition of Pair struct
  private struct Pair
  {
    public double First;
    public double Second;

    // Constructor
    public Pair(double first, double second)
    {
      First = first;
      Second = second;
    }
  }

  // Function to find the square root of
  // a complex number
  private static void ComplexRoot(int A, int B)
  {
    // Stores all the square roots
    var ans = new List<Pair>();

    // Stores the first square root
    double X1 = Math.Abs(Math.Sqrt((A + Math.Sqrt(A * A + B * B)) / 2));
    double Y1 = B / (2 * X1);

    // Push the square root in the ans
    ans.Add(new Pair(X1, Y1));

    // Stores the second square root
    double X2 = -1 * X1;
    double Y2 = B / (2 * X2);

    // If X2 is not 0
    if (X2 != 0)
    {
      // Push the square root in
      // the array ans[]
      ans.Add(new Pair(X2, Y2));
    }

    // Stores the third square root
    double X3 = (A - Math.Sqrt(A * A + B * B)) / 2;

    // If X3 is greater than 0
    if (X3 > 0)
    {
      X3 = Math.Abs(Math.Sqrt(X3));
      double Y3 = B / (2 * X3);

      // Push the square root in
      // the array ans[]
      ans.Add(new Pair(X3, Y3));

      // Stores the fourth square root
      double X4 = -1 * X3;
      double Y4 = B / (2 * X4);

      if (X4 != 0)
      {
        // Push the square root
        // in the array ans[]
        ans.Add(new Pair(X4, Y4));
      }
    }

    // Prints the square roots
    Console.WriteLine("The Square roots are: \n");

    foreach (var p in ans)
    {
      Console.Write($"{p.First:0.000000}");

      if (p.Second > 0)
      {
        Console.Write("+");
      }

      if (p.Second != 0)
      {
        Console.Write($"{p.Second:0.000000} * i\n");
      }
      else
      {
        Console.WriteLine();
      }
    }
  }

  // Driver code
  static void Main(string[] args)
  {
    int A = 0, B = 1;

    ComplexRoot(A, B);
  }
}

// This code is contributed by phasing17

JavaScript

<script>

// Javascript program for the above approach

// Function to find the square root of
// a complex number
function complexRoot(A, B)
{
    // Stores all the square roots
    var ans = [];

    // Stores the first square root
    var X1 = Math.abs(Math.sqrt((A + Math.sqrt(A * A
                                   + B * B))
                         / 2));
    var Y1 = B / (2 * X1);

    // Push the square root in the ans
    ans.push([X1, Y1]);

    // Stores the second square root
    var X2 = -1 * X1;
    var Y2 = B / (2 * X2);

    // If X2 is not 0
    if (X2 != 0) {

        // Push the square root in
        // the array ans[]
        ans.push([X2, Y2]);
    }

    // Stores the third square root
    var X3 = (A - Math.sqrt(A * A + B * B)) / 2;

    // If X3 is greater than 0
    if (X3 > 0) {
        X3 = Math.abs(Math.sqrt(X3));
        var Y3 = B / (2 * X3);

        // Push the square root in
        // the array ans[]
        ans.push([X3, Y3]);

        // Stores the fourth square root
        var X4 = -1 * X3;
        var Y4 = B / (2 * X4);

        if (X4 != 0) {

            // Push the square root
            // in the array ans[]
            ans.push([X4, Y4]);
        }
    }

    // Prints the square roots
    document.write( "The Square roots are: <br>");

    ans.forEach(p => {
        document.write( p[0].toFixed(6));
        if (p[1] > 0)
            document.write( "+");
        if (p[1])
            document.write( p[1].toFixed(6)
                 + "*i<br>" );
        else
            document.write("<br>");
    });
}

// Driver Code
var A = 0, B = 1;
complexRoot(A, B);

</script>

Output:

The Square roots are: 
0.707107+0.707107*i
-0.707107-0.707107*i

Time Complexity: O(1)
Auxiliary Space: O(1)

Square root of two Complex Numbers

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