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Math Help - rationalize a function

  1. #1
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    rationalize a function

    Show how the expression (-2i+1) is equivalent to (3-i)/(1+i)

    = (3-i)x(1-i) / (1+i)x(1-i)
    = (3i-3-i^2+1) / (i^2-1)

    What is next

    Thanks in advance
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  2. #2
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    We have

    1-2i=\underbrace{1-2i\cdot \frac{1+i}{3-i}}_{1}\cdot \frac{3-i}{1+i},

    now 1-2i\cdot \frac{1+i}{3-i}=\frac{1-i+2}{3-i}=1, and we're done.
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  3. #3
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    ???

    Well specifically the question asks "Which of the following expressions is equivalent to (3-i)/(1+i)" and then it gives you a choice of 4 expression (one being 1-2i .
    Do I have to solve each one or is there a quicker way???
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  4. #4
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    ???

    Also my friend said that you can plug (-1) into i^2 in my solution but I am not sure why
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  5. #5
    Senior Member Stroodle's Avatar
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    \frac{3-i}{1+i} =\frac{(3-i)(1-i)}{(1+i)(1-i)} =\frac{3-4i+i^2}{1-i^2} =\frac{2-4i}{2} =1-2i
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  6. #6
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    Quote Originally Posted by 05holtel View Post
    Also my friend said that you can plug (-1) into i^2 in my solution but I am not sure why
    Because i2= -1, of course.
    Sorry. "sup" is "HTML" for "superscript". It should be i^2= -1.
    Last edited by HallsofIvy; July 30th 2009 at 09:11 AM.
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  7. #7
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    I dont understand what i2= -1 means,
    Sorry
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  8. #8
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    HallsofIvy meant that i^2 = -1.


    01
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  9. #9
    Senior Member Stroodle's Avatar
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    Remember that i=\sqrt{-1} so i^2=(\sqrt{-1})^2=-1
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  10. #10
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    Is "i" always equal to the square root of -1
    Your patience is greatly appreciated [IMG]file:///C:/DOCUME%7E1/Elyse/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]
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  11. #11
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    Quote Originally Posted by Stroodle View Post
    Remember that i=\sqrt{-1} so i^2=(\sqrt{-1})^2=-1
    I wince when I see that. -1, like any complex number except 0, has two square roots. Unlike real numbers, since the complex numbers do not form and ordered field, we cannot just say "the positive number such that". I prefer just to assert that i^2= -1 and leave it at that. There are ways to define the complex numbers that avoid that problem.
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