Help multiplying complex numbers. I cant seem to figure out how to get the math part sorted out correctly. Can anyone help?
| xXxFarriexXx (10) |
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// s Complex Numbers.cpp : This program should allow the user to add, subtract, and multiply complex numbers.
//
#include "stdafx.h"
#include <iostream>
using namespace std;
class Complex
{
public:
void output()
{
cout << "Your complex number is: " << r << "+" << i << "i" << endl;
}
void input()
{
cout << "Enter the real part of the complex number: ";
cin >> r;
cout << "Enter the imaginary part of the complex number: ";
cin >> i;
}
void add (Complex n1, Complex n2)
{
r = n1.r + n2.r;
i = n1.i + n2.i;
}
void subtract( Complex n1, Complex n2)
{
r= n1.r- n2.r;
i = n1.i - n2.i;
}
void multiply (Complex n1, Complex n2)
{
n1.r = n1.r * n2.r + n1.r * n2.i;
n1.i = n1.i * n2.r + n1.i * n2.i ;
n2.r = n2.r*n1.r+ n1.r * n2.i;
n2.i = n2.i * n1.r+ n2.i * n1.i;
r = n1.r + n2.r;
i = n1.i + n2.i;
}
private:
int r;
int i;
};
int main()
{
Complex number1, number2, sum, differnce, product;
number1.input();
number2.input();
sum.add(number1, number2);
differnce.subtract(number1, number2);
product.multiply(number1, number2);
sum.output();
differnce.output();
product.output();
return 0;
}
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Last edited on
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| jumper007 (361) |
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Firstly, put your code in [code] [ /code ] tags...
Secondly, the "math" part of multiplying complex numbers is quite easy.
(x + yi)(a + bi) = (xa – yb) + (xb + ya)i |
I am not sure if I understood your question. So if this is not what you actually asked, please let me know.
Best of wishes,
~ Raul ~
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| xXxFarriexXx (10) |
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r = (n1.r * n2.r) - (n1.i * n2.i);
i = n1.r * n2.r + n1.i * n2.r ;
r + i * (i);
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Yes that right.. so if i type it into my program it would look like this correct?
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| jumper007 (361) |
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Yes. If by r + i * (i); you mean the output (cout), then yes.
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| xXxFarriexXx (10) |
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Nevermind I got it~ Thanks so much for the help its very much appreciated :)
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| jumper007 (361) |
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No problem. Glad I could help. :)
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