# How do you show that if n is a non-zero integer then (4/5+3/5i)^n != 1 ?

May 17, 2016

Consider the Real and imaginary parts of ${\left(4 + 3 i\right)}^{n}$ modulo $5$...

#### Explanation:

Consider the parallel sequences of integers defined recursively as follows:

${a}_{1} = 4$

${b}_{1} = 3$

${a}_{n + 1} = 4 {a}_{n} - 3 {b}_{n}$

${b}_{n + 1} = 4 {b}_{n} + 3 {a}_{n}$

Then modulo $5$ these sequences repeat every $4$ terms:

$a : 4 , 2 , 1 , 3 , 4 , 2 , 1 , 3 , \ldots$

$b : 3 , 4 , 2 , 1 , 3 , 4 , 2 , 1 , \ldots$

Now:

${\left(4 + 3 i\right)}^{n} = {a}_{n} + {b}_{n} i$ for $n = 1 , 2 , 3 , \ldots$

So:

$I m \left({\left(4 + 3 i\right)}^{n}\right) = {b}_{n} \ne 0$ for $n = 1 , 2 , 3 , \ldots$

and:

$I m \left({\left(\frac{4}{5} + \frac{3}{5} i\right)}^{n}\right) = {b}_{n} / {5}^{n} \ne 0$ for $n = 1 , 2 , 3 , \ldots$

So:

${\left(\frac{4}{5} + \frac{3}{5} i\right)}^{n} \ne 1$

When $n$ is a negative integer:

${\left(\frac{4}{5} + \frac{3}{5} i\right)}^{n} = \frac{1}{\frac{4}{5} + \frac{3}{5} i} ^ \left(- n\right) \ne 1$ too.

$\textcolor{w h i t e}{}$
Bonus

One corollary of this result is that the integer powers of $\left(\frac{4}{5} + \frac{3}{5} i\right)$ are dense in the unit circle.

To prove that, you can apply the pigeonhole principle to the unit circle subdivided into arbitrarily small arcs.

May 19, 2016

Here's an alternative proof using some properties of polynomials and algebraic numbers...

#### Explanation:

$\left(\frac{4}{5} + \frac{3}{5} i\right)$ is not the zero of a linear polynomial with integer coefficients. So its minimal polynomial over $\mathbb{Z}$ must be at least quadratic.

Note that:

$5 {\left(\frac{4}{5} + \frac{3}{5} i\right)}^{2} - 8 \left(\frac{4}{5} + \frac{3}{5} i\right) + 5 = 0$

and $5$ and $8$ are coprime.

Hence the minimum polynomial for $\left(\frac{4}{5} + \frac{3}{5} i\right)$ over $\mathbb{Z}$ is:

$5 {z}^{2} - 8 z + 5 = 0$

Since this is not monic, it cannot be a factor of ${z}^{n} - 1$ for any $n \ne 0$.

Hence: $\left(\frac{4}{5} + \frac{3}{5} i\right)$ is not an $n$th root of unity.