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Math Help - Complex number trig question.

  1. #1
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    Complex number trig question.

    Show that

    \frac{1+sin\theta+icos\theta}{1+sin\theta-icos\theta}=sin\theta+icos\theta

    and hence that

     <br />
(1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5 <br />
+i(1+sin\frac{\pi}{5}-icos\frac{\pi}{5})^5=0

    I have done the first part and am going round in circles trying to show the second part, I was trying to make use of de moivre's theorem but to no avail. Please help.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    EDIT: Oops, I misread and thought you were saying you were going in circles with the first part. Sorry
    Last edited by Unknown008; September 13th 2010 at 10:44 AM.
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  3. #3
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    Like I said, I have done the first part, just wasn't able to do the second, and would like help for that part, thanks.
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  4. #4
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    Hello,

    Trying to solve the second problem, I made a few manipulations, but I've seemed to arrive at a contradiction.

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+i(1+sin\f  rac{\pi}{5}-icos\frac{\pi}{5})^5=0

    Using i^5=i

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+(i+isin\f  rac{\pi}{5}+cos\frac{\pi}{5})^5=0

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})^5

    Using \sqrt[5]{-1}=-1

    1+sin\frac{\pi}{5}+icos\frac{\pi}{5}=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})

    1+sin\frac{\pi}{5}+icos\frac{\pi}{5}+i+isin\frac{\  pi}{5}+cos\frac{\pi}{5}=0

    1+sin\frac{\pi}{5}+cos\frac{\pi}{5} +i(1+sin\frac{\pi}{5}+cos\frac{\pi}{5})=0

    which only holds if the real and imaginary parts are equivalently 0. But because 1+sin\frac{\pi}{5}+cos\frac{\pi}{5} is not, it's weird. My calculator says that I'm wrong.
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  5. #5
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    Quote Originally Posted by Dfrtbx View Post
    Hello,

    Trying to solve the second problem, I made a few manipulations, but I've seemed to arrive at a contradiction.

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+i(1+sin\f  rac{\pi}{5}-icos\frac{\pi}{5})^5=0

    Using i^5=i

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+(i+isin\f  rac{\pi}{5}+cos\frac{\pi}{5})^5=0

    (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})^5

    Using \sqrt[5]{-1}=-1

    1+sin\frac{\pi}{5}+icos\frac{\pi}{5}=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5}) The contradiction comes here, where you are taking the fifth root of both sides of an equation. A complex number has five fifth roots, and you can't assume that you are getting the same one on both sides of the equation. To take a much simpler example, (–1)^2 = 1^2, but you can't take the square root of both sides and conclude that –1 = 1.

    1+sin\frac{\pi}{5}+icos\frac{\pi}{5}+i+isin\frac{\  pi}{5}+cos\frac{\pi}{5}=0

    1+sin\frac{\pi}{5}+cos\frac{\pi}{5} +i(1+sin\frac{\pi}{5}+cos\frac{\pi}{5})=0

    which only holds if the real and imaginary parts are equivalently 0. But because 1+sin\frac{\pi}{5}+cos\frac{\pi}{5} is not, it's weird. My calculator says that I'm wrong.
    To solve the second problem, notice that \sin\theta+i\cos\theta = i(\cos\theta-i\sin\theta) = i(\cos(-\theta) + i\sin(-\theta)). So \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=i(\cos(-\theta) + i\sin(-\theta)). Now put \theta = \pi/5, take the fifth power of both sides and use de Moivre's theorem.
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