Dados dos enteros positivos A y B que representan el número complejo Z en la forma Z = A + i * B , la tarea es encontrar la raíz cuadrada del número complejo dado.
Ejemplos:
Entrada: A = 0, B =1
Salida:
Las raíces cuadradas son:
0,707107 + 0,707107*i
-0,707107 – 0,707107*iEntrada: A = 4, B = 0
Salida:
Las raíces cuadradas son:
2
-2
Enfoque: El problema dado se puede resolver con base en las siguientes observaciones:
- Se sabe que la raíz cuadrada de un número complejo también es un número complejo.
- Entonces, considerando la raíz cuadrada del número complejo igual a X + i*Y , el valor de (A + i*B) se puede expresar como:
- A + i * B = (X + i * Y) * (X + i * Y)
- A + yo * segundo = X 2 – Y 2 + 2 * yo * X * Y
- Igualando el valor de partes reales y complejas individualmente:
A partir de las observaciones anteriores, calcule el valor de X e Y usando la fórmula anterior e imprima el valor (X + i*Y) como el valor resultante de la raíz cuadrada del número complejo dado.
A continuación se muestra la implementación del enfoque anterior:
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
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>
Producción:
The Square roots are: 0.707107+0.707107*i -0.707107-0.707107*i
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)