- #1

- 108

- 0

## Homework Statement

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.

## Homework 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?

## 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: