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Math Help - De Moivre's Theorem Question

  1. #1
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    De Moivre's Theorem Question

    Hi everyone!

    I have this question:

    If z = 1 + i√3, use de Moivre's theorem to find z^5 in cartesian form.

    I'm a bit lost, do I find all the roots?

    The steps I've taken so far are:
    1. Convert to polar form
    2. Find the roots.
    Do I then convert all of the roots back into cartesian?

    Cheers.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by falconed View Post
    Hi everyone!

    I have this question:

    If z = 1 + i√3, use de Moivre's theorem to find z^5 in cartesian form.

    I'm a bit lost, do I find all the roots?

    The steps I've taken so far are:
    1. Convert to polar form
    2. Find the roots.
    Do I then convert all of the roots back into cartesian?

    Cheers.
    yes

    what were the answers you got?
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  3. #3
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    Thanks for the quick reply!

    The answers I have are (they're still in polar form):

    z0 = 2^(1/5) . e^(pi/15)i
    z1 = 2^(1/5) . e^(7pi/15)i
    z2 = 2^(1/5) . e^(13pi/15)i
    z3 = 2^(1/5) . e^(19pi/15)i
    z4 = 2^(1/5) . e^(5pi/3)i

    So I just convert these back to cartesian and I'm done?

    I was under the impression that somehow I am supposed to get 1 answer. This method will give me 5 answers. :S

    Something like this makes more sense to me:

    z = 2.e^(pi/3)i
    z^5 = [2.e^(pi/3)i]^5
    = 32.e^(5pi/3)i
    = 32[cos(5pi/3) + i.sin(5pi/3)]
    = 16 - 16i√3
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by falconed View Post
    Thanks for the quick reply!

    The answers I have are (they're still in polar form):

    z0 = 2^(1/5) . e^(pi/15)i
    z1 = 2^(1/5) . e^(7pi/15)i
    z2 = 2^(1/5) . e^(13pi/15)i
    z3 = 2^(1/5) . e^(19pi/15)i
    z4 = 2^(1/5) . e^(5pi/3)i

    So I just convert these back to cartesian and I'm done?

    I was under the impression that somehow I am supposed to get 1 answer. This method will give me 5 answers. :S

    Something like this makes more sense to me:

    z = 2.e^(pi/3)i
    z^5 = [2.e^(pi/3)i]^5
    = 32.e^(5pi/3)i
    = 32[cos(5pi/3) + i.sin(5pi/3)]
    = 16 - 16i√3
    oh, i got confused by the question you asked. you mentioned "roots", in that case you should have 5 answers.

    if what the problem actually asked though, was to find z^5, then your last method is the one to choose. do not mix up roots with powers. if the question wanted roots, it would say "roots"
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  5. #5
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    Quote Originally Posted by falconed View Post
    Hi everyone!

    I have this question:

    If z = 1 + i√3, use de Moivre's theorem to find z^5 in cartesian form.

    I'm a bit lost, do I find all the roots?

    The steps I've taken so far are:
    1. Convert to polar form
    2. Find the roots.
    Do I then convert all of the roots back into cartesian?

    Cheers.
    Finding (fifth) roots would mean finding z^{1/5}. Is that what the question aks you to do ....?
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Finding (fifth) roots would mean finding z^{1/5}. Is that what the question aks you to do ....?
    Nah it says z^5.

    I got confused too, as I thought de Moivre's was only about finding the roots.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by falconed View Post
    Nah it says z^5.

    I got confused too, as I thought de Moivre's was only about finding the roots.
    then the last thing you did is what you want.

    de Moivre's theorem is not about finding roots. It is just useful in finding them. the theorem says:

    (\cos \theta + i \sin \theta)^n = \cos n \theta + i \sin n \theta for n any integer

    that's it. nothing about finding roots. it is just a way to simplify things
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