Question

1. [20 pts] Please write the functions addComplex, multiplyComplex, compareComplex, and printComplex, whose declarations are given as pseudocode below. All functions will operate on complex numbers, whose real and imaginary parts are given separately as double values. function addComplex input: real,:double, imgi: double, real2:double, img,:double output: realsum: double, imgsum double function multiplyComplex input: reall:double, img! : double, real2:double, output: realproduct:double, imgproduct :double img2 : double function compareComplex input: reali: double, imgi:double, real2:double, img2:double output: result:int where it is: 1 if complex1 complexz e if complex, complex2 -1 if complex, K complexz function printComplex input: real:double, img: double After writing the functions, test them with some example values so as to make sure that they are working as expected

question should be solved in C language

Answer 1

Hi,

The main challenge in the given problem is returning 2 values from a single function (the real part and the imaginary part).

This can be done in several ways.

1.You can use an array as an argument in the function and update the array with the returning values in the function itself.

2.You can pass pointers of the variables (as arguments)which are made to store the returning values.

3. You can create a structure and return a structure object (which indeed can have several data type values which can be updated by a function)

Here is the code for the array method approach-

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

void addComplex(double real1,double img1, double real2, double img2,double sum[])
{
   sum[0]=real1+real2;
   sum[1]=img1+img2;
  
}

void multiplyComplex(double real1,double img1, double real2, double img2,double prod[])
{
   prod[0]=real1*real2-img1*img2;
   prod[1]=real1*img2+img1*real2;
  
}

int compareComplex(double real1,double img1, double real2, double img2)
{
   double a=sqrt(pow(real1,2)+pow(img1,2));
   double b=sqrt(pow(real2,2)+pow(img2,2));
   if(a>b)
   {
       return 1;
   }
   else if(a==b)
   {
       return 0;
   }
   else{
       return -1;
   }  
}

void printComplex(double real1,double img1)
{
   printf("The complex number is : %lf + i*%lf \n",real1,img1);
}

int main()
{
   double real1,real2,img1,img2;
   double sum[2];
   double prod[2];
   int comp;
   printf("Enter the real part of 1st number\n");
   scanf("%lf",&real1);
   printf("Enter the complex part of 1st number\n");
   scanf("%lf",&img1);
   printf("Enter the real part of 2nd number\n");
   scanf("%lf",&real2);
   printf("Enter the complex part of 2nd number\n");
   scanf("%lf",&img2);
   addComplex(real1,img1,real2,img2,sum);
   printf("The result after adding is:\n");
   printComplex(sum[0],sum[1]);
   multiplyComplex(real1,img1,real2,img2,prod);
   printComplex(prod[0],prod[1]);
   comp=compareComplex(real1,img1,real2,img2);
   if(comp==1)
   {
       printf("complex1 > complex2\n");
   }
   else if(comp==0)
   {
       printf("complex1 = complex2\n");
   }
   else{
       printf("complex1 < complex2");
   }
  
}

A sample run of the code looks as follows-

Enter the real part of 1st number Enter the complex part of 1st number 4 Enter the real part of 2nd number Enter the complex part of 2nd number 2 The result after adding is: The complex number is: 4.000000 i*6.000000 The complex number is-5.000000 i*10.000000 complex1 > complex2 Process exited after 3.143 seconds with return value 0 Press any key to continue . . .

In the structure procedure, we basically create a structure called complex which has 2 double data types named real and img which represent the real and imaginary part of a real number.

Just like the array approach , we pass the real and img. part values into the function but this time we first create 2 instances of these structures (which represent our 2 complex numbers ) and pass their real and img. data fields as arguments.

In the function itself we create another structure which represents our result value. THis structure is then returned by the function.

The return value is collected by another structure defined in Main.

The code looks like this-

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

struct complex{
   double real;
   double img;
};
struct complex addComplex(double real1,double img1, double real2, double img2)
{
   struct complex sum;
   sum.real=real1+real2;
   sum.img=img1+img2;
   return sum;
  
}
struct complex multiplyComplex(double real1,double img1, double real2, double img2)
{
   struct complex prod;
   prod.real=real1*real2-img1*img2;
   prod.img=real1*img2+img1*real2;
   return prod;
  
}
int compareComplex(double real1,double img1, double real2, double img2)
{
   double a=sqrt(pow(real1,2)+pow(img1,2));
   double b=sqrt(pow(real2,2)+pow(img2,2));
   if(a>b)
   {
       return 1;
   }
   else if(a==b)
   {
       return 0;
   }
   else{
       return -1;
   }  
}
void printComplex(double real1,double img1)
{
   printf("The complex number is : %lf + i*%lf \n",real1,img1);
}
int main()
{
   struct complex c1,c2;
   struct complex r_sum;
   struct complex r_prod;
   int comp;
  
   printf("Enter the real part of 1st number\n");
   scanf("%lf",&c1.real);
   printf("Enter the complex part of 1st number\n");
   scanf("%lf",&c1.img);
   printf("Enter the real part of 2nd number\n");
   scanf("%lf",&c2.real);
   printf("Enter the complex part of 2nd number\n");
   scanf("%lf",&c2.img);
  
   r_sum=addComplex(c1.real,c1.img,c2.real,c2.img);
   r_prod=multiplyComplex(c1.real,c1.img,c2.real,c2.img);
   printf("The result after adding is:\n");
   printComplex(r_sum.real,r_sum.img);
   printf("The result after multiplication is:\n");
   printComplex(r_prod.real,r_prod.img);
  
   comp=compareComplex(c1.real,c1.img,c2.real,c2.img);
   if(comp==1)
   {
       printf("complex1 > complex2\n");
   }
   else if(comp==0)
   {
       printf("complex1 = complex2\n");
   }
   else{
       printf("complex1 < complex2");
   }
  
  
}

A sample output of the code looks like-

Enter the real part of 1st number 1 Enter the complex part of 1st number 2 Enter the real part of 2nd number Enter the complex part of 2nd number 4 The result after adding is: The complex number is 4.0000001*6.000000 The result after multiplication is: The complex number is -5.000000 i*10.000000 complex1 complex2 Process exited after 2.934 seconds with return value 19 Press any key to continue . . .