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Math Help - i operator help

  1. #1
    Junior Member Jarod_C's Avatar
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    i operator help

    I think I should make sure I have this right so I don't screw up the rest of my homework. Here's what I've got,

    (j3)(-j7)(-j)(j)=(j^2)(-j^2)(21)=(-1)(1)(21)=-21

    So, am I right or where did I go wrong. I think it's whether or not I signed that right. Thanks for the help, Jarod.
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  2. #2
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    Re: i operator help

    Quote Originally Posted by Jarod_C View Post
    I think I should make sure I have this right so I don't screw up the rest of my homework. Here's what I've got,

    (j3)(-j7)(-j)(j)=(j^2)(-j^2)(21)=(-1)(1)(21)=-21

    So, am I right or where did I go wrong. I think it's whether or not I signed that right. Thanks for the help, Jarod.
    Are you using j as most people use i~?

    Does (j3) mean j^3\text{ or }3j~?
    Thanks from Jarod_C
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  3. #3
    Junior Member Jarod_C's Avatar
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    Re: i operator help

    Sorry I should have clarified that. This is for the use of electronics. Since "i" is used to represent current we use j instead. Also j3 means j\cdot 3 or 3j Sorry for the confusion.
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  4. #4
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    Re: i operator help

    Quote Originally Posted by Jarod_C View Post
    Sorry I should have clarified that. This is for the use of electronics. Since "i" is used to represent current we use j instead. Also j3 means j\cdot 3 or 3j Sorry for the confusion.
    In that case (j3)(-j7)(-j)(j)=(3)(7)(j^3)=-21j

    \\j^0=1\\j^1=j\\j^2=-1\\j^3=-j

    j^N=j^{N\text{Mod}~4}
    Thanks from Jarod_C
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  5. #5
    Junior Member Jarod_C's Avatar
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    Re: i operator help

    Thanks, as usual the help here is always fast and courteous.
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  6. #6
    Junior Member Jarod_C's Avatar
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    Re: i operator help

    Quick question, why does this result in  j^3 instead of  j^4 ? That's how I figured my answer of -21 because  j^4=(j^2)(j^2)=(-1)(-1)=1

    One more thing, do I need to make a to thread for a new problem? I thought I read that somewhere once.
    Last edited by Jarod_C; February 3rd 2014 at 04:48 PM.
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