# Modular Exponentiation of Complex Numbers

Given four integers A, B, K, M. The task is to find (A + iB)K % M which is a complex number too. A + iB represents a complex number.

Examples:

Input : A = 2, B = 3, K = 4, M = 5
Output: 1 + i*0

Input : A = 7, B = 3, K = 10, M = 97
Output: 25 + i*29

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisite: Modular Exponentiation

Approach:
An efficient approach is similar to the modular exponentiation of a single number. Here, instead of a single we have two number A, B. So, pass a pair of integers as a parameter to the function instead of a single number.

Below is the implementation of the above approach :

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Function to multiply two complex numbers modulo M ` `pair<``int``, ``int``> Multiply (pair<``int``, ``int``> p, pair<``int``, ``int``> q, ` `                                                    ``int` `M) ` `{ ` `    ``// Multiplication of two complex numbers is  ` `    ``// (a + ib)(c + id) = (ac - bd) + i(ad + bc) ` `     `  `    ``int` `x = ((p.first * q.first) % M - (p.second *  ` `                                    ``q.second) % M + M) % M; ` `     `  `    ``int` `y = ((p.first * q.second) % M + (p.second *  ` `                                          ``q.first) % M) %M; ` ` `  `    ``// Return the multiplied value ` `    ``return` `{x, y}; ` `} ` ` `  ` `  `// Function to calculate the complex modular exponentiation ` `pair<``int``, ``int``> compPow(pair<``int``, ``int``> complex, ``int` `k, ``int` `M) ` `{ ` `    ``// Here, res is initialised to (1 + i0) ` `    ``pair<``int``, ``int``> res = { 1, 0 };  ` `     `  `    ``while` `(k > 0)  ` `    ``{ ` `        ``// If k is odd ` `        ``if` `(k & 1) ` `        ``{ ` `            ``// Multiply 'complex' with 'res' ` `            ``res = Multiply(res, complex, M);  ` `        ``} ` `         `  `        ``// Make complex as complex*complex ` `        ``complex = Multiply(complex, complex, M); ` `         `  `        ``// Make k as k/2 ` `        ``k = k >> 1;  ` `    ``} ` `     `  `    ``//Return the required answer ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `A = 7, B = 3, k = 10, M = 97; ` `     `  `    ``// Function call ` `    ``pair<``int``, ``int``> ans = compPow({A, B}, k, M); ` `     `  `    ``cout << ans.first << ``" + i"` `<< ans.second;     ` `     `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `static` `class` `pair  ` `{  ` `    ``int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to multiply two complex numbers modulo M ` `static` `pair Multiply (pair p, pair q, ``int` `M) ` `{ ` `    ``// Multiplication of two complex numbers is  ` `    ``// (a + ib)(c + id) = (ac - bd) + i(ad + bc) ` `     `  `    ``int` `x = ((p.first * q.first) % M - ` `             ``(p.second * q.second) % M + M) % M; ` `     `  `    ``int` `y = ((p.first * q.second) % M +  ` `             ``(p.second * q.first) % M) % M; ` ` `  `    ``// Return the multiplied value ` `    ``return` `new` `pair(x, y); ` `} ` ` `  ` `  `// Function to calculate the  ` `// complex modular exponentiation ` `static` `pair compPow(pair complex, ``int` `k, ``int` `M) ` `{ ` `    ``// Here, res is initialised to (1 + i0) ` `    ``pair res = ``new` `pair(``1``, ``0` `);  ` `     `  `    ``while` `(k > ``0``)  ` `    ``{ ` `        ``// If k is odd ` `        ``if` `(k % ``2` `== ``1``) ` `        ``{ ` `            ``// Multiply 'complex' with 'res' ` `            ``res = Multiply(res, complex, M);  ` `        ``} ` `         `  `        ``// Make complex as complex*complex ` `        ``complex = Multiply(complex, complex, M); ` `         `  `        ``// Make k as k/2 ` `        ``k = k >> ``1``;  ` `    ``} ` `     `  `    ``// Return the required answer ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `A = ``7``, B = ``3``, k = ``10``, M = ``97``; ` `     `  `    ``// Function call ` `    ``pair ans = compPow(``new` `pair(A, B), k, M); ` `     `  `    ``System.out.println(ans.first + ``" + i"` `+  ` `                       ``ans.second);  ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to multiply two complex numbers modulo M ` `def` `Multiply (p, q, M): ` `     `  `    ``# Multiplication of two complex numbers is  ` `    ``# (a + ib)(c + id) = (ac - bd) + i(ad + bc) ` `    ``x ``=` `((p[``0``] ``*` `q[``0``]) ``%` `M ``-` `\ ` `         ``(p[``1``] ``*` `q[``1``]) ``%` `M ``+` `M) ``%` `M ` `     `  `    ``y ``=` `((p[``0``] ``*` `q[``1``]) ``%` `M ``+` `\ ` `         ``(p[``1``] ``*` `q[``0``]) ``%` `M) ``%``M ` ` `  `    ``# Return the multiplied value ` `    ``return` `[x, y] ` ` `  `# Function to calculate the ` `# complex modular exponentiation ` `def` `compPow(``complex``, k, M): ` `     `  `    ``# Here, res is initialised to (1 + i0) ` `    ``res ``=` `[``1``, ``0``]  ` `     `  `    ``while` `(k > ``0``): ` `         `  `        ``# If k is odd ` `        ``if` `(k & ``1``): ` `             `  `            ``# Multiply 'complex' with 'res' ` `            ``res ``=` `Multiply(res, ``complex``, M) ` `         `  `        ``# Make complex as complex*complex ` `        ``complex` `=` `Multiply(``complex``, ``complex``, M) ` `         `  `        ``# Make k as k/2 ` `        ``k ``=` `k >> ``1` `     `  `    ``# Return the required answer ` `    ``return` `res ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``A ``=` `7` `    ``B ``=` `3` `    ``k ``=` `10` `    ``M ``=` `97` `     `  `    ``# Function call ` `    ``ans ``=` `compPow([A, B], k, M) ` `     `  `    ``print``(ans[``0``], ``"+ i"``, end ``=` `"") ` `    ``print``(ans[``1``]) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `public` `class` `pair  ` `{  ` `    ``public` `int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to multiply two complex numbers modulo M ` `static` `pair Multiply (pair p, pair q, ``int` `M) ` `{ ` `    ``// Multiplication of two complex numbers is  ` `    ``// (a + ib)(c + id) = (ac - bd) + i(ad + bc) ` `     `  `    ``int` `x = ((p.first * q.first) % M - ` `             ``(p.second * q.second) % M + M) % M; ` `     `  `    ``int` `y = ((p.first * q.second) % M +  ` `             ``(p.second * q.first) % M) % M; ` ` `  `    ``// Return the multiplied value ` `    ``return` `new` `pair(x, y); ` `} ` ` `  ` `  `// Function to calculate the  ` `// complex modular exponentiation ` `static` `pair compPow(pair complex, ``int` `k, ``int` `M) ` `{ ` `    ``// Here, res is initialised to (1 + i0) ` `    ``pair res = ``new` `pair(1, 0 );  ` `     `  `    ``while` `(k > 0)  ` `    ``{ ` `        ``// If k is odd ` `        ``if` `(k % 2 == 1) ` `        ``{ ` `            ``// Multiply 'complex' with 'res' ` `            ``res = Multiply(res, complex, M);  ` `        ``} ` `         `  `        ``// Make complex as complex*complex ` `        ``complex = Multiply(complex, complex, M); ` `         `  `        ``// Make k as k/2 ` `        ``k = k >> 1;  ` `    ``} ` `     `  `    ``// Return the required answer ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `A = 7, B = 3, k = 10, M = 97; ` `     `  `    ``// Function call ` `    ``pair ans = compPow(``new` `pair(A, B), k, M); ` `     `  `    ``Console.WriteLine(ans.first + ``" + i"` `+  ` `                      ``ans.second);  ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```25 + i29
```

Time complexity: O(log k).

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