# Rotating two different complex numbers

1. Jul 7, 2010

### daniel_i_l

Let's say that I have two complex numbers, a and b, with different arguments. From a few "experiments" with a computer, I think that there always exists a positive integer n such that -pi/2 <= Arg(a^n) <= pi/2 and pi/2 <= Arg(b^n) <= 3pi/2.
In other words, if Arg(a) = thetaA and Arg(b) = thetaB, then there exists an n such that
-pi/2 <= n*thetaA <= pi/2 and pi/2 <= n*thetaB <= 3pi/2 up to a multiple of 2pi.
Is this true? If so, how can I start to prove it?
Thanks,
Daniel

2. Jul 7, 2010

### Filip Larsen

It's not true for a = -1 and b = 1. In general, if both a and b have rational arguments they will only have a finite number of integer power images so it seems likely that you may find a whole range of argument pairs for which your condition will never be satisfied.

3. Jul 7, 2010

### tmccullough

Do you just need them to be in different halves? In other words, is it sufficient that
$$|\textrm{arg}(a^n)-\textrm{arg}(b^n)| = n\cdot |\textrm{arg}(a)-\textrm{arg}(b)| \geq \pi$$
-modulo $2\pi$ of course?

As Flip Larsen commented, you can't necessarily pick which is in which half...the above is definitely true, though. The proof is pretty much immediate.