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

Question

2007 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T23:15:05

Consider the Real and imaginary parts of #(4+3i)^n# modulo #5#...

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 #5# these sequences repeat every #4# terms:

#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 #n# is a negative integer:

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

#color(white)()#
Bonus

One corollary of this result is that the integer powers of #(4/5+3/5i)# are dense in the unit circle.

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

Answer 2 · 2016-05-19T18:06:22

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

#(4/5+3/5i)# is not the zero of a linear polynomial with integer coefficients. So its minimal polynomial over #ZZ# must be at least quadratic.

Note that:

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

and #5# and #8# are coprime.

Hence the minimum polynomial for #(4/5+3/5i)# over #ZZ# is:

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

Since this is not monic, it cannot be a factor of #z^n-1# for any #n != 0#.

Hence: #(4/5+3/5i)# is not an #n#th root of unity.