Related Articles

# Modular Exponentiation of Complex Numbers

• Difficulty Level : Expert
• Last Updated : 23 Aug, 2019

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).

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up