1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  4. Apr 2, 2010 #3


    User Avatar
    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


    User Avatar
    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
    \left (\frac{\sqrt{3}}{2} + \frac{1}{2}i \right )^{n}
    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


    User Avatar
    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Newton binomial problem