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++ 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 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 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 |
<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)