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Math Help - complex numer express (1+i)^9/(1-i)^9

  1. #1
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    complex numer express (1+i)^9/(1-i)^9

    complex numer express (1+i)^9/(1-i)^9-img_20140603_093007-1-.jpgcomplex numer express (1+i)^9/(1-i)^9-img_20140603_093120-1-.jpgi express (1+i)^9 as 16surd9(cos (9pi/4) +i sin (9pi/4)) and (1-i)^9 as 16surd9(cos (-9pi/4) -i sin (9pi/4)) , then i divide both, but my ans is (1+i)/(1-i) , which is differnt with the ans given, why cant i do in this way? i have attached the sample ans
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  2. #2
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    Re: complex numer express (1+i)^9/(1-i)^9

    Quote Originally Posted by delso View Post
    why cant i do in this way? i have attached the sample ans
    Whoever wrote that did not know much about the fundamentals.

    Here is a very useful fact: $\dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}$

    Thus $\dfrac{{1 + i}}{{1 - i}} = \dfrac{{{{(1 + i)}^2}}}{2} = i$ and $i^9=i$.
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  3. #3
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    Re: complex numer express (1+i)^9/(1-i)^9

    [QUOTE=delso;821295]Click image for larger version. 

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ID:	31060i express (1+i)^9 as 16surd9(cos (9pi/4) +i sin (9pi/4)) and (1-i)^9 as 16surd9(cos (-9pi/4) -i sin (9pi/4))[quote]
    This is wrong. |1+ i|= \sqrt{2} and (\sqrt{2})^9= \sqrt{2}(\sqrt{2})^8= \sqrt{2}(2)^4= 16\sqrt{2}.
    I don't know what you mean by 'surd9' but I'm sure you did not mean " \sqrt{2}"!

    , then i divide both, but my ans is (1+i)/(1-i) , which is differnt with the ans given, why cant i do in this way? i have attached the sample ans
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  4. #4
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    Re: complex numer express (1+i)^9/(1-i)^9

    Hello, delso!

    \text{(b) Evaluate: }\:\left(\frac{1+i}{1-i}\right)^9

    Multiply by \frac{1+i}{1+i}

    . . \left(\frac{1+i}{1-i}\cdot{\color{red}\frac{1+i}{1+i}}\right)^9 \;=\;\left(\frac{1+2i-1}{1-(\text{-}1)}\right)^9 \;=\;\left(\frac{2i}{2}\right)^9 \;=\;i^9 \;=\;i
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