# Newton binomial problem

1. Apr 2, 2010

### Taturana

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Considering the binomial theorem as:

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

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:

$$\binom{n}{1} \left( \frac{\sqrt{3}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \binom{n}{3} \left( \frac{\sqrt{3}}{2} \right )^{n-3} \left ( \frac{1}{2}i \right )^{3}$$

Is that right?

3. The attempt at a solution

So I tried to solve that equation as the follows:

$$\frac{n!}{\left( n-1 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-1} \left ( \frac{1}{2}i \right ) = \frac{n!}{3!\left( n-3 \right )!} \left( \frac{3^{\frac{1}{2}}}{2} \right )^{n-3} \left ( \frac{1}{2} \right )^{3}\left(-i \right )$$

$$\frac{n!}{\left( n-1 \right )!} \ \ \frac{3^{\frac{n-1}{2}}}{2^{n-1}} \ \ \frac{i}{2} = \frac{n!}{3!\left( n-3 \right )!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n-3}} \ \ \left ( \frac{-i}{2^{3}} \right )$$

$$n \ \ \frac{3^{\frac{n-1}{2}}}{2^{n}} \ \ \ i = \frac{n\left(n-1 \right )\left(n-2 \right )}{3!} \ \ \frac{3^{\frac{n-3}{2}}}{2^{n}} \ \ \ \left( -i \right)$$

$$n \ \ 3^{\frac{n-1}{2}} = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{\frac{n-3}{2}}$$

$$n = - \frac{n\left(n-1 \right )\left(n-2 \right )}{6} \ \ 3^{-1}$$

$$18 = - \left(n-1 \right )\left(n-2 \right )$$

$$18 = - \left( n^{2} -3n +2 \right )$$

and finally

$$-n^{2} + 3n + 16 = 0$$

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

Thank you

Last edited: Apr 2, 2010
2. Apr 2, 2010

### pseudogenius

I don't know but try this:

$$\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{n}= p$$

$$\frac{\sqrt{3}}{2}+\frac{1}{2}i= p^{\frac{1}{n}}$$

$$\sqrt{3}+i= 2p^{\frac{1}{n}}$$

$$\left(\sqrt{3}+i\right)^{n}= 2^{n}p$$

$$2^{n}$$ times a positive real number is a positive real number, so find a positive real solution to the simpler equation

$$\left(\sqrt{3}+i\right)^{n}$$

3. Apr 2, 2010

### lanedance

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

re-writing in that form should simplify the problem as when you multiply by itself you get $( r e^{i \theta})^2 = r^2 e^{2 i \theta}$ as $r$ is a real numer, so is $r^2$

Last edited: Apr 2, 2010
4. Apr 3, 2010

### Taturana

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?

5. Apr 3, 2010

### Dick

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.

6. Apr 3, 2010

### Taturana

I consider the complex as $r e^{i \theta}$ and do what after that?