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

##### 2 Answers

Consider the Real and imaginary parts of

#### Explanation:

Consider the parallel sequences of integers defined recursively as follows:

#a_1 = 4#

#b_1 = 3#

#a_(n+1) = 4a_n-3b_n#

#b_(n+1) = 4b_n+3a_n#

Then modulo

#a: 4, 2, 1, 3, 4, 2, 1, 3,...#

#b: 3, 4, 2, 1, 3, 4, 2, 1,...#

Now:

#(4+3i)^n = a_n + b_n i# for#n=1,2,3,...#

So:

#Im((4+3i)^n) = b_n != 0# for#n = 1,2,3,...#

and:

#Im((4/5+3/5i)^n) = b_n/5^n != 0# for#n=1,2,3,...#

So:

#(4/5+3/5i)^n != 1#

When

#(4/5+3/5i)^n = 1/(4/5+3/5i)^(-n) != 1# too.

**Bonus**

One corollary of this result is that the integer powers of

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

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

#### Explanation:

Note that:

#5(4/5+3/5i)^2-8(4/5+3/5i)+5 = 0#

and

Hence the minimum polynomial for

#5z^2-8z+5 = 0#

Since this is not monic, it cannot be a factor of

Hence: