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Math Help - Raising complex numbers to a power

  1. #1
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    Raising complex numbers to a power

    Hi,

    I'm using j when I know some textbooks use i, I'm not sure what the difference is if there is any.

    I have the question (4j^5 - 5j^4 + 2j^3 -3j^2)^2 and it's supposed to evaluate to -8j, I just don't remember how to get that answer. To me it seems that something like 4j^10 could be considered as 4j^2 since there are 5 pairs of -1 resulting from multiplying j by j, so 4j^10 would be 4(-1). That does not get me the right answer though. What is the flaw in my thinking?

    Thanks!
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  2. #2
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    Re: Raising complex numbers to a power

    No there is no difference. The standard convention is to use "i" (for imaginary number), but some texts use "j" because "i" is being used for another purpose (I know this is the convention used by electrical engineers).

    Anyway, to help you with your problem, recall that j^2 = -1, which means j^3 = -j, j^4 = 1, and then they repeat. That should simplify the expression greatly before trying to square.
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    Re: Raising complex numbers to a power

    Quote Originally Posted by KevinShaughnessy View Post
    Hi,

    I'm using j when I know some textbooks use i, I'm not sure what the difference is if there is any.

    I have the question (4j^5 - 5j^4 + 2j^3 -3j^2)^2 and it's supposed to evaluate to -8j, I just don't remember how to get that answer. To me it seems that something like 4j^10 could be considered as 4j^2 since there are 5 pairs of -1 resulting from multiplying j by j, so 4j^10 would be 4(-1). That does not get me the right answer though. What is the flaw in my thinking?

    Thanks!
    Usually j is used in electricity where i is for the electric current.

    Conxtruct a table to yourself like this

    j^0 = j^4 = j^8 =j^(4*n) = 1 n is integer

    Continue the table that starts with
    j^1 =
    j^2 =
    j^3 =

    use that table to find the answer to your problem
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    Re: Raising complex numbers to a power

    I'm afraid I'm still stuck. When I break down (4j^5 - 5j^4 + 2j^3 -3j^2)^2, j^5 is the same as j, so 4j^5 can be rewritten as 4j. 5j^4 is the same as 1, so that can be rewritten as 5(1). 2j^3 is the same as negative j, so that can be rewritten as or -2j. j^2 is the same as negative one, so that can be rewritten as +3. That leaves me with (4j-5-2j+3)^2 which leads to 16j^2 -25 -4j^2 + 9 or -28.
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    Re: Raising complex numbers to a power

    Simplify before squaring: (4j-5-2j+3)^2 = (2j-2)^2 or be more careful when multiplying out before simplifying: (4j-5-2j+3)(4j-5-2j+3) = 4j(4j-5-2j+3) - 5(4j-5-2j+3) -2j(4j-5-2j+3)+3(4j-5-2j+3). You missed a LOT of terms.
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    Re: Raising complex numbers to a power

    Quote Originally Posted by SlipEternal View Post
    Simplify before squaring: (4j-5-2j+3)^2 = (2j-2)^2 or be more careful when multiplying out before simplifying: (4j-5-2j+3)(4j-5-2j+3) = 4j(4j-5-2j+3) - 5(4j-5-2j+3) -2j(4j-5-2j+3)+3(4j-5-2j+3). You missed a LOT of terms.
    Ah, I see where I went wrong. For some reason it didn't even occur to me that you needed to do that. Thanks a lot for your help!
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