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Math Help - DeMoivre's Theorem

  1. #1
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    DeMoivre's Theorem

    4. b) Use De Moivre's theorem to express (-square root 2 / 2 + i square root 2 / 2 )^50 in the form a + bi

    I found out on the unit circle the point is at 135 degrees. Once applying the exponent rules the point becomes 270 degrees (0,-1).

    At this point I am stuck trying to convert to a + bi form.

    I believe it is radius 2^25 so we get: 2^25 (0 - 1i) for an answer of 2^25i. Yes?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thekrown View Post
    4. b) Use De Moivre's theorem to express (-square root 2 / 2 + i square root 2 / 2 )^50 in the form a + bi

    I found out on the unit circle the point is at 135 degrees. Once applying the exponent rules the point becomes 270 degrees (0,-1).

    At this point I am stuck trying to convert to a + bi form.

    I believe it is radius 2^25 so we get: 2^25 (0 - 1i) for an answer of 2^25i. Yes?
    Rewrite the expression as \left(\cos\!\left(\tfrac{3\pi}{4}\right)+i\sin\!\l  eft(\tfrac{3\pi}{4}\right)\right)^{50}.

    Applying DeMoivre's Theorem, we get

    \cos\!\left(\tfrac{75}{2}\pi\right)+i\sin\!\left(\  tfrac{75}{2}\pi\right).

    Due to the periodic behavior of sine and cosine, this is equivalent to \cos\!\left(\tfrac{3\pi}{2}\right)+i\sin\!\left(\t  frac{3\pi}{2}\right)=-i

    Does this make sense?
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  3. #3
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    Isn't the radius 2^25?

    I don't see where cos 75 pi / 2 and i sin 75 pi / 2 come from.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thekrown View Post
    Isn't the radius 2^25?
    No, its 1 because \left|\cos\!\left(\tfrac{3\pi}{4}\right)+i\sin\!\l  eft(\tfrac{3\pi}{4}\right)\right|=\sqrt{\cos^2\!\l  eft(\tfrac{3\pi}{4}\right)+\sin^2\!\left(\tfrac{3\  pi}{4}\right)}=\sqrt{1}=1.

    I don't see where cos 75 pi / 2 and i sin 75 pi / 2 come from.
    By DeMoivre's Theorem,

    \begin{aligned}\left(\cos\!\left(\tfrac{3\pi}{4}\r  ight)+i\sin\!\left(\tfrac{3\pi}{4}\right)\right)^{  50}&=\cos\!\left(\tfrac{3\pi}{4}\cdot 50\right)+i\sin\!\left(\tfrac{3\pi}{4}\cdot 50\right)\\ &=\cos\!\left(\tfrac{150\pi}{4}\right)+i\sin\!\lef  t(\tfrac{150\pi}{4}\right)\\ &=\cos\!\left(\tfrac{75\pi}{2}\right)+i\sin\!\left  (\tfrac{75\pi}{2}\right)\end{aligned}
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  5. #5
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    Okay I understand where the 3 pi / 2 comes from, it is 270 degrees. I had the wrong radius, I think I was using the initial radius maybe?

    I believe I made the mistake of not using the formula to find the radius. I'm having trouble with this but I'll give it another shot.

    If I have trouble I will let you know. Thanks.
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  6. #6
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    On the unit circle the point is at 0,-1 for 270 degrees or cos 3 pi / 2 + sin 3 pi / 2, yes?

    If so can I use the formula r = square root of a^2 + b^2 which is the same as square root 0^2 + (-1)^2 = 1?
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