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Newton binomial problem

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the minor value of the natural number n such that [tex]\left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}[/tex] be a real positive number.
    EDIT: n must not be 0.

    2. Relevant equations

    Considering the binomial theorem as:

    [tex]{\left(x+y\right)}^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^k\,\![/tex]

    I realized that we need to cancel the terms that we have i and -i. So we need that the terms where k = 1 and k = 3 be opposite. So we need that:

    [tex]\binom{n}{1} \left( \frac{\sqrt{3}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \binom{n}{3} \left( \frac{\sqrt{3}}{2} \right )^{n-3} \left ( \frac{1}{2}i \right )^{3}[/tex]

    Is that right?

    3. The attempt at a solution

    So I tried to solve that equation as the follows:

    [tex]\frac{n!}{\left( n-1 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \frac{n!}{3!\left( n-3 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-3} \left ( \frac{1}{2} \right )^{3}\left(-i \right )[/tex]

    [tex]\frac{n!}{\left( n-1 \right )!} \ \ \frac{3^{\frac{n-1}{2}}}{2^{n-1}} \ \ \frac{i}{2} = \frac{n!}{3!\left( n-3 \right )!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n-3}} \ \ \left ( \frac{-i}{2^{3}} \right )[/tex]

    [tex]n \ \ \frac{3^{\frac{n-1}{2}}}{2^{n}} \ \ \ i = \frac{n\left(n-1 \right )\left(n-2 \right )}{3!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n}} \ \ \ \left( -i \right)[/tex]

    [tex]n \ \ 3^{\frac{n-1}{2}} = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{\frac{n-3}{2}}[/tex]

    [tex]n = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{-1}[/tex]

    [tex]18 = - \left(n-1 \right )\left(n-2 \right )[/tex]

    [tex]18 = - \left( n^{2} -3n +2 \right )[/tex]

    and finally

    [tex]-n^{2} + 3n + 16 = 0[/tex]

    But the answers to that quadratic equation aren't natural numbers. What did I do wrong?

    Thank you
     
    Last edited: Apr 2, 2010
  2. jcsd
  3. Apr 2, 2010 #2
    I don't know but try this:

    [tex]\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{n}= p[/tex]

    [tex]\frac{\sqrt{3}}{2}+\frac{1}{2}i= p^{\frac{1}{n}}[/tex]

    [tex]\sqrt{3}+i= 2p^{\frac{1}{n}}[/tex]

    [tex]\left(\sqrt{3}+i\right)^{n}= 2^{n}p[/tex]


    [tex]2^{n}[/tex] times a positive real number is a positive real number, so find a positive real solution to the simpler equation

    [tex]\left(\sqrt{3}+i\right)^{n}[/tex]
     
  4. Apr 2, 2010 #3

    lanedance

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    Homework Helper

    follwoing on from psuedogenius, any complex number [itex] a+ bi [/itex] can be written [itex] r e^{i \theta} [/itex]

    re-writing in that form should simplify the problem as when you multiply by itself you get [itex]( r e^{i \theta})^2 = r^2 e^{2 i \theta} [/itex] as [itex]r[/itex] is a real numer, so is [itex]r^2[/itex]
     
    Last edited: Apr 2, 2010
  5. Apr 3, 2010 #4
    I can't solve this way because we didn't learn that on the classes. But thank you for the help.

    Does anyone have another suggestions? What did I do wrong on my try?
     
  6. Apr 3, 2010 #5

    Dick

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    Science Advisor
    Homework Helper

    What you did wrong in your try is equating the k=1 and k=3. For one thing that doesn't make them cancel. For another thing, even it did, what about k=5, 7, etc? For another thing you also want the result to be positive, not just real. If you don't know polar form, then why not just try evaluating
    [tex]
    \left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}
    [/tex]
    for successive values of n, until you get a value that lets you see the answer.
     
  7. Apr 3, 2010 #6
    Thank you for the reply.

    I know the polar form: z = |z| * (cos a + i * sin a). But how do I solve the problem as lanedance said? I don't understand...

    I consider the complex as [itex]
    r e^{i \theta}
    [/itex] and do what after that?
     
  8. Apr 3, 2010 #7

    Dick

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    Science Advisor
    Homework Helper

    Try to figure out what theta is. Then (r*exp(i*theta))^n=r^n*exp(i*n*theta). That's real and positive if n*theta is a multiple of 2*pi, right?
     
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