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Complex Number once agn.

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Given that [tex](1-\sqrt{3}i)^{n}[/tex] is real and positive , use de Moivre's Theorem to show that the values of n are terms of arithmetic progression.


    2. Relevant equations



    3. The attempt at a solution

    I've worked it out . n=3k s.t k element of positive integers.
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 22, 2010 #2

    Mark44

    Staff: Mentor

    Are you sure you have given us the problem exactly as stated? My guess is that they're asking you to show that the values of (1 - sqrt(3)i)n are terms of an arithmetic progression. As you've stated the problem it doesn't make much sense.
     
  4. Feb 22, 2010 #3
    Hi Mark! I've checked with the question again , and its exactly what i've stated.
     
  5. Feb 22, 2010 #4

    Mark44

    Staff: Mentor

    OK - just checking. Apparently the problem asks you to find which values of n are such that (1 - sqrt(3)i)^n are real and positive. What is 1 - sqrt(3)i in polar form? That should help you figure out a lot more easily what the powers of 1 - sqrt(3)i look like.
     
  6. Feb 22, 2010 #5
    yeap. i've worked it out in my attempted solution and i'm convinced that it is a arithmetic progression as n has 0 as first term and 3 as common difference for each term.
     
  7. Feb 23, 2010 #6

    Mentallic

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    Not quite.

    The question is asking for when it's real and positive. Since 2n is always positive for natural n, then you are searching for when the complex part has an argument of 0. i.e. [itex]\theta=2k\pi[/itex] where k is all integers.

    You can even see for yourself that for n=3 the value of [tex]\left(1-\sqrt{3}i\right)^3[/tex] is real, but negative.
     
  8. Feb 23, 2010 #7

    HallsofIvy

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    You still don't understand what Mark44 was asking.

    In your original post, you gave a complex number to the "n" power and asked to "show that the values of n are terms of arithmetic progression."

    Well, that's trivial! The values of n are 1, 2, 3, 4, 5, ..., with common difference 1!
     
  9. Feb 23, 2010 #8

    Mark44

    Staff: Mentor

    1, 2, 3, ... don't work in the problem as I understand it. IMO the problem is poorly stated and misleading. A better restatement of the problem might be "For what n is the expression (1 - sqrt(3)i)n real and positive? Show that these values of n form an arithmetic progression."
     
  10. Feb 24, 2010 #9

    Mentallic

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    I believe the question was clear enough. It first mentions that [tex](1-\sqrt{3})^n[/tex] is real and positive, which obviously indicates that n is not all natural numbers. Then it goes on to say show that the values of n are in arithmetic progression.
    Shouldn't be confusing whatsoever in my opinion.
     
  11. Feb 24, 2010 #10

    Mark44

    Staff: Mentor

    I still maintain that the problem could have been clearer if rephrased. "Given that (1 - i*sqrt(3))^n is real and positive, ..." would cause one to reasonably infer that this statement was true for integer values of n, or possibly just nonnegative integer values.
     
  12. Feb 25, 2010 #11

    Mentallic

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    Maybe for someone without a basic knowledge of complex numbers and exponentials, but this is untrue for everyone that has posted in this thread.

    Also, I believe that realizing the base case of natural numbers n=1 contradicts the statement that [itex](1-i\sqrt{3})^n[/itex] is real and positive would be easier than having to make some reasonable judgement on what the question is saying.

    ... I stand by my statement that it's obvious what the question was asking.
     
  13. Feb 25, 2010 #12

    HallsofIvy

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    Ah, thanks to both Mark44 and Mentallic. I completely misunderstood the problem. It might be better phrased "show that those n such that [itex](1- i\sqrt{3})^n[/itex] are real and positive form an arithmetic progression".
     
  14. Feb 25, 2010 #13

    Mark44

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    I rest my case.:biggrin:
     
  15. Feb 25, 2010 #14

    Mentallic

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    I guess I was wrong...

    Haha just kidding :wink:
     
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