Does there exists a $k\in\mathbb{Z}$ such that $|a^{k}z-1|\geq\sqrt{2}$ for given $1\neq a\in S^{1}$ and $z\in S^{1}$?

Question

Suppose that $a$ and $z$ are complex numbers of modulus $1$ with $a\neq1$. Can we always find an integer $k\in\mathbb{Z}$ such that $|a^{k}z-1|\geq\sqrt{2}$? My intuition says yes. I think it suffices to prove that we can find $k\in\mathbb{Z}$ such that $a^{k}z$ lies in the left-hand half of the unit circle. In this case I can draw a picture that sort of convinces me.

Answer 1

hint

Put $$a=e^{i\alpha}\text{ and }z=e^{i\beta}$$

for $ k\in \Bbb Z,$

$$a^kz-1=$$ $$(\cos(k\alpha+\beta)-1)+i\sin(k\alpha+\beta)$$ and $$|a^kz-1|=2|\cos(\frac{k\alpha+\beta}{2})|$$

So you want $$|\cos(\frac{k\alpha+\beta)}{2})|\ge \frac{\sqrt{2}}{2}$$ $$(=\cos(\frac{\pi}{4}))$$