Write a program to find sum of two complex numbers?

Answer 1

#include<iostream.h> #include<conio.h> class complex { int a,b; public: void read() { cout<<"\n\nEnter the REAL PART : "; cin>>a; cout<<"\n\nEnter the IMAGINARY PART : "; cin>>b; } complex operator +(complex c2) { complex c3; c3.a=a+c2.a; c3.b=b+c2.b; return c3; } complex operator -(complex c2) { complex c3; c3.a=a-c2.a; c3.b=b-c2.b; return c3; } complex operator *(complex c2) { complex c3; c3.a=(a*c2.a)-(b*c2.b); c3.b=(b*c2.a)+(a*c2.b); return c3; } complex operator /(complex c2) { complex c3; c3.a=((a*c2.a)+(b*c2.b))/((c2.a*c2.a)+(c2.b*c2.b)); c3.b=((b*c2.a)-(a*c2.b))/((c2.a*c2.a)+(c2.b*c2.b)); return c3; } void display() { cout<<a<<"+"<<b<<"i"; } }; void main() { complex c1,c2,c3; int choice,cont; do { clrscr(); cout<<"\t\tCOMPLEX NUMBERS\n\n1.ADDITION\n\n2.SUBTRACTION\n\n3.MULTIPLICATION\n\n4.DIVISION"; cout<<"\n\nEnter your choice : "; cin>>choice; if(choice==1choice==2choice==3choice==4) { cout<<"\n\nEnter the First Complex Number"; c1.read(); cout<<"\n\nEnter the Second Complex Number"; c2.read(); } switch(choice) { case 1 : c3=c1+c2; cout<<"\n\nSUM = "; c3.display(); break; case 2 : c3=c1-c2; cout<<"\n\nResult = "; c3.display(); break; case 3 : c3=c1*c2; cout<<"\n\nPRODUCT = "; c3.display(); break; case 4 : c3=c1/c2; cout<<"\n\nQOUTIENT = "; c3.display(); break; default : cout<<"\n\nUndefined Choice"; } cout<<"\n\nDo You Want to Continue?(1-Y,0-N)"; cin>>cont; }while(cont==1); getch(); }