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Math Help - Imaginary unit

  1. #1
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    Imaginary unit

    Why does i equal the square root of -1 and i^2 equal -1?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by HSS8 View Post
    Why does i equal the square root of -1
    that is simply how it is defined.

    and i^2 equal -1?
    well, squaring gets rid of the square root.

    see here and here
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  3. #3
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    Quote Originally Posted by HSS8 View Post
    Why does i equal the square root of -1 and i^2 equal -1?
    As the previous reply correctly states, this is how "the imaginary" is defined. Math folks needed a number to do a certain thing, so they said "let's create that number, and call it `i\mbox{'}, since clearly it's 'imaginary' and can't have any possible 'real world' use or application." (This is, by the way, similar to how negative numbers got their start.)

    There is a lot of math, science, and engineering that couldn't happen nowadays, had this number not been invented.

    When you study complex numbers (numbers with the i in them or involving taking the square root of negative values), make sure you remember the new rule: Always take care of the negative inside the square root first, before doing any other computations or simplifications!

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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by HSS8 View Post
    Why does i equal the square root of -1 and i^2 equal -1?
    The complex numbers are constructed so that i^2=-1, but there are 2 square roots of -1 and i by construction is one of them.

    (note I describe the complex numbers as constructed as they are built to have the properties needed so that certain algebraic process that we need make sense.)

    CB
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  5. #5
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    The "construction" CaptainBlack refers to is this: Define the complex numbers to be the set of pairs of real numbers (a, b) with addition defined by (a,b)+ (c,d)= (a+c,b+d), (a,b)*(c,d)= (ac-bd,ad+bc). Since (a,0)+ (c,0)= (a+c,0) and (a,0)*(c,0)= (ac-0(0),a(0)+0(c))= (ac,0), we can identify the complex numbers of the form (a,0) with the real numbers. Now, (0,1)*(0,1)= (0(0)-1(1),0(1)+1(0))= (-1, 0) which we associate with -1.

    If we represent (0,1) by i, and, as I said, (1,0) by 1, we have (a,b)= a(1,0)+ b(0,1)= a+ bi. And, then, i^2= (0,1)*(0,1)= (-1,0)= i.
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