//  Given two sets of complex numbers, the first set contains 3 numbers and the
//  second set contains 2 numbers.thsi program outputs all the possible imaginary numbers that  are the multiplications of
//  all complex numbers in the first set with those in the second. Note that the
//  size of this set will be 6.
//  Author:rahule@cse.iitk.ac.in

#include<stdio.h>
main()
{
  double a[2][3],b[2][2],real,img;
  int i,j;
  printf("\nfist set");
  printf("\n----------");
  for(i=0;i<3;i++)	//read the first set of imaginary numbers in 'a'
  {
  printf("\nenter the real part of the  number  :");
  scanf("%lf",&a[0][i]);
  printf("\nenter the imaginary part of the number :");
  scanf("%lf",&a[1][i]);
  }
  printf("\nthe first set has been entered\n");
  
  printf("\nsecond set");
  printf("\n----------");
   for(i=0;i<2;i++)		//read the second set of imaginary numbers in set 'b'
  {
  printf("\nenter the real part of the  number  :");
  scanf("%lf",&b[0][i]);
  printf("\nenter the imaginary part of the number :");
  scanf("%lf",&b[1][i]);
  }
  printf("\nthe second set has been entered\n");
  printf("\nthe all possible products are ,");
  
  for(i=0;i<3;i++)
    for(j=0;j<2;j++)
    {
      real=a[0][i]*b[0][j];		//multiply ith number in set a with jth number in ser b,and store in real,img
      img=a[1][i]*b[1][j];
      
      //now prints 'real + i img'
      //note that if imaginary part is negative,we should print "real - i img",NOT 'real i-img'
      printf("\n(%lf",a[0][i]);
      if(a[1][i]<0)
	printf(" - i %lf)",-(a[1][i]));
      else
	printf(" + i %lf)",a[1][i]);
      
      printf("*(%lf",b[0][j]);
      if(b[1][j]<0)
	printf(" - i %lf)",-(b[1][j]));
      else
	printf(" + i %lf)",b[1][j]);
      
      printf(" = (%lf",real);
      if(img<0)
	printf(" - i %lf)",-(img));
      else
	printf(" + i %lf)",img);
     }
     printf("\n");
     
}