Step by Step solution about How to resolve the algorithm Roots of unity step by step in the C programming language

This C program generates a table of complex nth roots of unity for various values of n and i, where n is the order of the root and i is the index of the root.

• A complex number is represented in rectangular form in the program. A complex number z can be written as z = c + si, where c and s are real numbers, and i is the imaginary unit (i^2 = -1).
• The program uses the atan2 function from the math.h library to calculate the value of PI2, which is 8 times the arctangent of 1, or approximately 2 * pi.
• The main loop of the program iterates through n values from 1 to 9 and i values from 0 to n-1 for each n.
• For each pair of n and i, the program calculates the values of c and s based on the conditions given:
  • If i is 0, c is set to 1.
  • If n is 4 times i, s is set to 1.
  • If n is 2 times i, c is set to -1.
  • If 3 times n is 4 times i, s is set to -1.
  • Otherwise, c and s are calculated using cos and sin.
• The program then prints the values of c and s in the specified format:
  • c is printed if it is non-zero.
  • s is printed with a leading "+" if it is positive, a leading "-" if it is negative, or with no leading sign if it is zero. The imaginary unit "i" is appended to s if its value is 1 or -1.
• The program prints a newline character after each row of the table.

Here is an example of the output produced by the program:

1
1,  -i,  -1,  -i,  1
1,         1,  -1,         -1,  1
1,  -1,  -i,         1,  -i
1,         i,  -1,         -i,  1
1,         -1,  -i,         i,  1
1,  -i,  -1,  -i,  1,  -1
1,         1,  -1,  -1,  1,  1
1,  -1,  -i,         1,  -i,  1
1,         i,  -1,         -i,  1,  1
1,         -1,  -i,         i,  1,  1

#include <stdio.h>
#include <math.h>

int main()
{
	double a, c, s, PI2 = atan2(1, 1) * 8;
	int n, i;

	for (n = 1; n < 10; n++) for (i = 0; i < n; i++) {
		c = s = 0;
		if (!i )		c =  1;
		else if(n == 4 * i)	s =  1;
		else if(n == 2 * i)	c = -1;
		else if(3 * n == 4 * i)	s = -1;
		else
			a = i * PI2 / n, c = cos(a), s = sin(a);

		if (c) printf("%.2g", c);
		printf(s == 1 ? "i" : s == -1 ? "-i" : s ? "%+.2gi" : "", s);
		printf(i == n - 1 ?"\n":",  ");
	}

	return 0;
}