The Problem: Suppose $f(z) = e^{i\theta}z$. Show that if $\theta$ is not a rational multiple of $\pi$, then the orbit of $ z \in \mathbb{C}$ is dense in the circle with radius $|z|$ and at the center of the origin.
My Attempt: Now I think that I am supposed to show that the orbit of z is dense inside the circle with radius z. Let $\delta > 0$ I can define another point $z_{0} \in \mathbb{C}$ which $0<|z_{0}|< |z|$. Then for any integer k, $|f^{k} (z_{0}) -f(z)| < \delta$. The problem I am having is choosing a $\delta$ so that the orbit of $z_{0}$ is contained inside the circle with radius z.
I am not also sure if I am on the right track.
Could you just give me some hints or feedback on what I need to do?
Thanks for all of your help!
I don't think you're thinking about this problem correctly. Note that $\left|f(z)\right| = \left|z\right|$, so that repeated applications of $f$ stay on the circle, not "in" ( meaning inside ) the circle. Also, I don't think the inequality you gave is correct.
Your function $f$ is just rotation by $\theta$ degrees. So, one way to think of the problem is rephrasing as: starting from the point $(1,0)$, can you successively rotate by $\theta$ to get a point on the circle with angle less than $\epsilon > 0$?
EDIT:
You need to start with a point $w$ on the circle, i.e. $|w| = |z|$ and then show that for any $\epsilon > 0$ there is a $k$ such that $|f^k(z) - w| < \epsilon$. This problem can be pretty easily reduced to showing that for any $\epsilon > 0$ there is a $k$ such that the angle between $f^k(z)$ and $w$ is less than $\epsilon$. So, let $\xi = Arg(z)$, so that $Arg(f^k(z)) = \xi + k\cdot \theta \pmod {2\pi}$, and let $\zeta = Arg(w)$. So, this reduces to showing that if $\theta$ is not a rational multiple of $\pi$, then there is a $k$ such that $\zeta - \xi - k\cdot \theta \pmod{2\pi}$ is less than $\epsilon$. I guess I'm being a little hand-wavy on that since there's no order on elements of $\mathbb{R}/2\pi\mathbb{R}$, but I think it's clear what I mean. Dividing everything by $2\pi$, this boils down to showing that given a real number $a$ between $0$ and $1$ and an irrational number $c$, there exists a natural number $k$ such that $a-k\cdot c \pmod 1$ is less than $\epsilon$.
Does this help?