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Program for power of a complex number in O(log n) in C++
Given a complex number in the form of x+yi and an integer n; the task is calculate and print the value of the complex number if we power the complex number by n.
What is a complex number?
A complex number is number which can be written in the form of a + bi, where a and b are the real numbers and i is the solution of the equation or we can say an imaginary number. So, simply putting it we can say that complex number is a combination of Real number and imaginary number.
Raising power of a complex number
To raise the power of a complex number we use the below formula −
(a+bi) (c+di)=( ac−bd )+(ad+bc )i
Like we have a complex number
2+3i and raising its power by 5 then we will get −
(2+3 i)5=(2+3 i)(2+3i)(2+3 i)(2+3 i)(2+3i)
Using the above formula we will get answer −
Example
Input: x[0] = 10, x[1] = 11 /*Where x[0] is the first real number and 11 is the second real number*/ n = 4 Output: -47959 + i(9240) Input: x[0] = 2, x[1] =3 n = 5 Output: 122 + i(597)
Approach we are using to solve the above problem −
So, the problem can be solved using iterative method easily but the complexity will be O(n), but we have to solve the problem in O(log n) time. For that we can −
- First take the input in form of an array.
- In function Power the x^n
- Check if n is 1, then return x
- Recursively call power pass x and n/2 and store its result in a variable sq.
- Check if dividing n by 2 leaves a remainder 0; if so then return the results obtained from cmul(sq, sq)
- Check if dividing n by 2 does not leaves a remainder 0; if so then return the results obtained from cmul(x, cmul(sq, sq)).
- In function cmul().
- Check if x1 = a+bi and x2 = x+di, then x1 * x2 = (a*c–b*d)+(b*c+d*a)i.
- Return and printthe results obtained.
Algorithm
Start Step 1-> declare function to calculate the product of two complex numbers long long* complex(long long* part1, long long* part2) Declare long long* ans = new long long[2] Set ans[0] = (part1[0] * part2[0]) - (part1[1] * part2[1]) Se ans[1] = (part1[1] * part2[0]) + part1[0] * part2[1] return ans Step 2-> declare function to return the complex number raised to the power n long long* power(long long* x, long long n) Declare long long* temp = new long long[2] IF n = 0 Set temp[0] = 0 Set temp[1] = 0 return temp End IF n = 1 return x End Declare long long* part = power(x, n / 2) IF n % 2 = 0 return complex(part, part) End return complex(x, complex(part, part)) Step 3 -> In main() Declare int n Declare and set long long* x = new long long[2] Set x[0] = 10 Set x[1] = -11 Set n = 4 Call long long* a = power(x, n) Stop
Example
#include <bits/stdc++.h>
using namespace std;
//calculate product of two complex numbers
long long* complex(long long* part1, long long* part2) {
long long* ans = new long long[2];
ans[0] = (part1[0] * part2[0]) - (part1[1] * part2[1]);
ans[1] = (part1[1] * part2[0]) + part1[0] * part2[1];
return ans;
}
// Function to return the complex number raised to the power n
long long* power(long long* x, long long n) {
long long* temp = new long long[2];
if (n == 0) {
temp[0] = 0;
temp[1] = 0;
return temp;
}
if (n == 1)
return x;
long long* part = power(x, n / 2);
if (n % 2 == 0)
return complex(part, part);
return complex(x, complex(part, part));
}
int main() {
int n;
long long* x = new long long[2];
x[0] = 10;
x[1] = -11;
n = 4;
long long* a = power(x, n);
cout << a[0] << " + i ( " << a[1] << " )" << endl;
return 0;
}Output
power of complex number in O(Log n) : -47959 + i ( 9240 )
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