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All the powers are added they dont make 10

  1. Nov 3, 2008 #1
    i would like to know why or how the following comes about.

    when j^10 =(j^4)^2j^2=1^2(-1)=1

    why from ^10 does a J^4 squared and a j^2 when all the powers are added they dont make 10.

    could you tell me a little about it.
     
  2. jcsd
  3. Nov 3, 2008 #2

    CompuChip

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    Re: simplyify

    Oh, but they do. Remember that (j^4)^2 = (j^4) (j^4).

    On the other hand, 1^2 x (-1) is not 1 :tongue:
     
  4. Nov 3, 2008 #3
    Re: simplyify

    so it is about making the powers add up,this is good im a second year science student but i could not remember the maths on this one ,im working though a book to get to know this kinda stuff.thank you for your help.
     
  5. Nov 3, 2008 #4

    CompuChip

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    Re: simplyify

    Actually you should remember these rules:
    • an * am = am + n
    • (an)m = am*n
    (for a real) for the rest of your life. If you forget them, you can go back to the case where n and m are integer numbers and write it out, for example:
    [tex]a^3 \times a^4 = (a \times a \times a) \times (a \times a \times a \times a) = (a \times a \times a \times a \times a \times a \times a) = a^7[/tex]
    whereas
    [tex](a^3)^4 = (a \times a \times a)^4 = (a \times a \times a) \times (a \times a \times a \times a) \times a \times a \times a) \times (a \times a \times a \times a) = (a \times a \times a \times a \times a \times a \times a \times a \times a \times a \times a \times a \times a \times a) = a^{12}[/tex]
     
  6. Nov 4, 2008 #5
    Re: simplyify

    hello again ive another one for you.

    on the equation (4-j3)^2 =16-j24-9

    i can see were the 16 and the 9 comes from but were does the j24
     
  7. Nov 4, 2008 #6
    Re: simplyify

    multiply it out,

    (4 - j3) (4 - j3)

    you'll see that there'll be a -j24 from the multiplication of the real number and the complex number.
     
  8. Nov 4, 2008 #7
    Re: simplyify

    well i did 4*4+-3*4+-3*-3+3*-3 but i got 25

    is there some thing wrong with my results.is the -j24 as were the j is a number=-1
     
  9. Nov 4, 2008 #8
    Re: simplyify

    j is a representation of the square root of -1, think of it as a variable.
    i dont think you did the multiplication right.

    (a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2
     
  10. Nov 5, 2008 #9
    Re: simplyify

    no but i think im getting used to it thank you for your help.
     
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