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Math Help - Demonstration envolving Complex numbers!! Help please

  1. #1
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    Unhappy Demonstration envolving Complex numbers!! Help please

    Please someone help me.. I've an exam in 2 days ... The following problem makes me nuts.. I can't get it right..

    Show that:

    cos(∏ - alpha) + i cos (∏/2 - alpha)
    -------------------------------------------- = cis (∏ - 2alpha)
    cis (alpha)

    I'm desperate... I've tried everything I know.. any help is appreciated
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  2. #2
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    Re: Demonstration envolving Complex numbers!! Help please

    Quote Originally Posted by josepbigorra View Post
    Please someone help me.. I've an exam in 2 days ... The following problem makes me nuts.. I can't get it right..

    Show that:

    cos(∏ - alpha) + i cos (∏/2 - alpha)
    -------------------------------------------- = cis (∏ - 2alpha)
    cis (alpha)

    I'm desperate... I've tried everything I know.. any help is appreciated
    What have you tried? It's hard to give you advice when we don't know what you've tried.

    I'd favor expanding out cos( \pi - \alpha ) = cos(\pi)~cos(\alpha) + sin(\pi)~sin(\alpha) = -cos(\alpha) etc.

    -Dan
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    Re: Demonstration envolving Complex numbers!! Help please

    What I've tried is the following:


    Then i proceed to the substitution in the expression from above, but I don't know how to continue.
    Thanks in advance
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    Re: Demonstration envolving Complex numbers!! Help please

    Quote Originally Posted by topsquark View Post
    What have you tried? It's hard to give you advice when we don't know what you've tried.

    I'd favor expanding out cos( \pi - \alpha ) = cos(\pi)~cos(\alpha) + sin(\pi)~sin(\alpha) = -cos(\alpha) etc.

    -Dan
    What I've tried is the following:


    Then i proceed to the substitution in the expression from above, but I don't know how to continue.
    Thanks in advance
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Demonstration envolving Complex numbers!! Help please

    Okay, you've got the cos( \pi - \alpha ) and sin( \pi / 2 - \alpha ) expressions. So you need to evaluate:
    \frac{-cos( \alpha ) + i ~sin( \alpha ) }{cos( \alpha ) + i~sin(\alpha)}

    You could do the division but recall Euler's theorem: cos(x) + i~sin(x) = e^{ix}. See what you can do with this.

    -Dan
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    Re: Demonstration envolving Complex numbers!! Help please

    Or "rationalize the denominator" by multiplying numerator and denominator by the conjugate of the denominator:
    \left(\frac{-cos(\alpha)+ isin(\alpha)}{cos(\alpha)+ isin(\alpha)}\right)\left(\frac{cos(\alpha)- isin(\alpha)}{cos(\alpha)- isin(\alpha)}\right)
    = -cos^2(\alpha)+ sin^2(\alpha)+ 2sin(\alpha)cos(\alpha).

    Now, what about that left side? cis(\pi- 2\alpha)= cos(\pi- 2\alpha)+ isin(\pi- 2\alpha)= -cos(2\alpha)+ isin(2\alpha)
    So what are cos(2\alpha) and sin(2\alpha)?
    Thanks from topsquark
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  7. #7
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    Re: Demonstration envolving Complex numbers!! Help please

    Quote Originally Posted by HallsofIvy View Post
    Or "rationalize the denominator" by multiplying numerator and denominator by the conjugate of the denominator:
    \left(\frac{-cos(\alpha)+ isin(\alpha)}{cos(\alpha)+ isin(\alpha)}\right)\left(\frac{cos(\alpha)- isin(\alpha)}{cos(\alpha)- isin(\alpha)}\right)
    = -cos^2(\alpha)+ sin^2(\alpha)+ 2sin(\alpha)cos(\alpha).

    Now, what about that left side? cis(\pi- 2\alpha)= cos(\pi- 2\alpha)+ isin(\pi- 2\alpha)= -cos(2\alpha)+ isin(2\alpha)
    So what are cos(2\alpha) and sin(2\alpha)?

    My friend thanks a lot for the info... I managed to get the problem working.. I really appreciate your help
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