# 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

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`

## Javascript

 `function` `pair(first, second) {` `    ``this``.first = first;` `    ``this``.second = second;` `}`   `function` `multiply(p, q, M) {` `   ``// Multiplication of two complex numbers is ` `    ``// (a + ib)(c + id) = (ac - bd) + i(ad + bc)` `    ``let x = ((p.first * q.first) % M - (p.second * q.second) % M + M) % M;` `    ``let y = ((p.first * q.second) % M + (p.second * q.first) % M) % M;` `    ``return` `new` `pair(x, y);` `}`   `function` `compPow(complex, k, M) {` `    ``let res = ``new` `pair(1, 0);` `    ``while` `(k > 0) {` `        ``if` `(k % 2 === 1) {` `            ``res = multiply(res, complex, M);` `        ``}` `        ``complex = multiply(complex, complex, M);` `        ``k = k >> 1;` `    ``}` `    ``return` `res;` `}`   `let A = 7, B = 3, k = 10, M = 97;` `let ans = compPow(``new` `pair(A, B), k, M);` `console.log(ans.first + ``" + i"` `+ ans.second);`   `// This code is contributed by abn95knd1.`

Output:

`25 + i29`

Time complexity: O(log k).

Auxiliary Space: O(1)

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