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Math Help - Another i operator

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

    Here's the problem,

    (1-i)(1-i)^2 so I worked it like this,

    (1-i)(1^2-i^2)=
    (1-i)(1-(-1))=
    (1-i)(2)=
    (2i)

    Does that look correct, or should I have started it by working the square in the second set of parentheses like this (1-i)(1-i)(1-i)
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  2. #2
    MHF Contributor
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    Re: Another i operator

    the second way.

    (x+iy)^2 = (x^2 - y^2) + i2xy

    NOT

    x^2 + (iy)^2
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  3. #3
    MHF Contributor

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    Re: Another i operator

    Quote Originally Posted by Jarod_C View Post
    Here's the problem,

    (1-i)(1-i)^2 so I worked it like this,

    (1-i)(1^2-i^2)=
    ??? Surely, you know that (a- b)^2 is NOT a^2- b^2!
    (a- b)^2= a^2- 2ab+ b^2. With a= 1, b= i, that is 1- 2(1)(i)+ (i)^2= 1- 2i- 1= -2i

    (1-i)(1-(-1))=
    (1-i)(2)=
    (2i)
    and certainly, (1- i)(2)= 2- 2i not "2i"!

    Does that look correct, or should I have started it by working the square in the second set of parentheses like this (1-i)(1-i)(1-i)
    It's not a matter of where you start, you need to review multiplication of complex numbers completely!
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