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Math Help - complexe2

  1. #1
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    complexe2

    Hello i have 3 exercices i finished them and i need the last questions

    Write in trigonometrical form the complex numbers:
    [1+itan(alfa)]^4
    ------------
    [1-itan(alfa)]

    it's one number but i don't know how to use maths letters here

    Thanks
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  2. #2
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    Quote Originally Posted by iceman1 View Post
    Hello i have 3 exercices i finished them and i need the last questions

    Write in trigonometrical form the complex numbers:
    [1+itan(alfa)]^4
    ------------
    [1-itan(alfa)]

    it's one number but i don't know how to use maths letters here

    Thanks
    There is a "clever" way to start simplifying this, but let's just take things at face value here.

    The first thing you want to do is to get rid of that "i" in the denominator:
    \frac{(1 + i~tan(\alpha))^4}{1 - i~tan(\alpha)} = \frac{(1 + i~tan(\alpha))^4}{1 - i~tan(\alpha)} \cdot \frac{1 + i~tan(\alpha)}{1 + i~tan(\alpha)}

    = \frac{(1 + i~tan(\alpha))^5}{1 - i^2~tan^2(\alpha)} = \frac{(1 + i~tan(\alpha))^5}{1 + tan^2(\alpha)} = \frac{(1 + i~tan(\alpha))^5}{sec^2(\alpha)}

    Now expand the numerator using the binomial theorem (or if you are masochistic, expand it out term by term) and separate the real and imaginary parts.

    -Dan
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  3. #3
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    Recall that \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }} so that \frac{1}{{1 - i\tan (\alpha )}} = \frac{{1 + i\tan (\alpha )}}{{\sec ^2 (\alpha )}}.

    \frac{{\left( {1 + i\tan (\alpha )} \right)^4 }}{{1 - i\tan (\alpha )}} = \frac{{\left( {1 + i\tan (\alpha )} \right)^5 }}{{\sec ^2 (\alpha )}}

    \frac{{\left( {1 + i\tan (\alpha )} \right)^5 }}{{\sec ^2 (\alpha )}} = \frac{{\frac{{\left( {\cos (\alpha ) + i\sin (\alpha )} \right)^5 }}{{\cos ^5 (\alpha )}}}}{{\sec ^2 (\alpha )}} = \frac{{\cos (5\alpha ) + i\sin (5\alpha )}}{{\cos ^3 (\alpha )}}
    Last edited by Plato; September 27th 2007 at 05:54 AM.
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  4. #4
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    This is a postscript, resulting from a PM.
    It seems that the problem is actually \left[ {\frac{{1 + \tan (\alpha )}}{{1 - \tan (\alpha )}}} \right]^4 .
    In which case;
    \begin{array}{rcl}<br />
 \left[ {\frac{{1 + i\tan (\alpha )}}{{1 - i\tan (\alpha )}}} \right]^4  & = & \left[ {\frac{{\cos (\alpha ) + i\sin (\alpha )}}{{\cos (\alpha ) - i\sin (\alpha )}}} \right]^4  \\ <br />
   & = & \left[ {\cos (2\alpha ) + i\sin (2\alpha )} \right]^4  \\ <br />
  & = & \cos (8\alpha ) + i\sin (8\alpha ) \\ <br />
 \end{array}
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