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.