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*iInput: 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:
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 numbervoid 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 Codeint main(){ int A = 0, B = 1; complexRoot(A, B); return 0;} |
Java
// Java program for the above approachimport 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 numberstatic 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 Codepublic 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 approachfrom math import sqrt# Function to find the square root of# a complex numberdef 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 Codeif __name__ == '__main__': A,B = 0, 1 complexRoot(A, B)# This code is contributed by mohit kumar 29 |
C#
// C# code to implement the approachusing 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 numberfunction 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 Codevar 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)
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