imaginary numbers

This is a discussion on imaginary numbers within the C++ Programming forums, part of the General Programming Boards category; You don't need to actually use complex numbers, you just need to know how they work and be clever with ...

  1. #16
    Pursuing knowledge confuted's Avatar
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    You don't need to actually use complex numbers, you just need to know how they work and be clever with some radical simplification.

    The Quadratic Formula. The quadratic equation ax^2+bx+c has the solutions

    x=(-b +- sqr (b^2-4ac))/2a
    i is the square root of -1. The part of the quadratic equation that could make complex numbers necessary is (sqr(b^2-4ac)). If (b^2-4ac) < 0 then you're going to need some complex numbers. For the example I'm about to give, let a=8, b=0, and c=2.

    b^2-4ac
    0^2-4(8)(2)
    -64
    At this point, we're trying to find the square root of -64.
    sqr(-64)==sqr(-1*64)==sqr(-1)*sqr(64)==8*sqr(-1)=8i

    In your code, if you just take the absolute value of b^2-4ac, take the square root of that, and slap an i after it (if b^2-4ac evaluates to <0) you'll be fine. Then, just stick that number into the quadratic formula...to finish it off, we have this:

    x=(-b +- sqr (b^2-4ac))/2a
    x=(-0+-8i)/(2*8)
    x=(+-8i)/16
    x=8i/16 or -8i/16
    x=.5i or -.5i

    See, never needed to program anything with complex numbers.
    Away.

  2. #17
    Senior Member joshdick's Avatar
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    First of all, I'd like to say that the solution proposed by Blackrat is IMO the best for your problem. Secondly, there is a complex number class in the standard header files. I believe it's complex.h, but I may be wrong. If you can't find that library or have trouble using it, write your own class for complex numbers. I did that for a class once. Not difficult at all if you know how to work with complex numbers. The only data members you need are double a, b;

    Hmmm, I never know how to end these posts. Umm, good luck!
    FAQ

    "The computer programmer is a creator of universes for which he alone is responsible. Universes of virtually unlimited complexity can be created in the form of computer programs." -- Joseph Weizenbaum.

    "If you cannot grok the overall structure of a program while taking a shower, you are not ready to code it." -- Richard Pattis.

  3. #18
    ¡Amo fútbol!
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    Originally posted by Gil22
    alright i think i actually need to use complex numbers......is therea way to declare these?
    Yeah, include complex and declare a std::complex. A lot of what you will need has already been implemented for you.

  4. #19
    Registered User
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    an easier way to find ou if a number is imaginary is to find out if the equation under the radical is negative. If so then output an i after the output of the answer. If not then just evaluate it regullarly. A little bit of the code would look like this:
    Code:
    float determinant;
    float root;
    
    determinant = b*b - (4*a*c);
    
    if(determinant < 0)
    {
            determinant = determinant * -1;
            root1 =  (b + Sqrt(determinant) )/ 2;
            root 2 = (b - Sqrt(determinant) )/ 2;
           
            cout << "The answers are: " << root1 << "i, " << root2 << "i." << endl;
    }
    
    else
    {
            root1 =  (b + Sqrt(determinant) )/ 2;
            root 2 = (b - Sqrt(determinant) )/ 2;
    
            cout << "The answers are: " << root1 << ", " << root2 << endl;
    }
    There you go. Hope it helps you out.

    -Bill
    cout << My_Balls;
    cin >> Jo_Mouf;

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