No geodesic ray has bounded distance to the logarithmic spiral

Let $c \colon (0,\infty) \to \mathbb{C}$, $c(r)=re^{i\log(r)}$ be a logarithmic spiral. I have shown $$|r-r'|\leq|c(r)-c(r')|\leq \sqrt{2}|r-r'|.$$ So $c$ is a quasi-isometry. Now I want to show that there is no geodesic ray $$f \colon [0,\infty) \to \mathbb{C},\, f(t) = te^{i \phi}+z $$ that has bounded distance to $c$.

Two mappings $f,g \colon D \to \mathbb{C}$ have bounded distance_ if $$ \sup_{x \in D} |f(x)-g(x)| < \infty $$ holds.

However, I can't show that for every $\phi$ and $z$ the term $$ |c(t)-f(t)|=|t(e^{i\log(t)}-e^{i \phi})-z| $$ is unbounded.

Does anyone have an idea or a hint?

The background of the question is that one has a stability of quasi-geodesics in $\delta$-hyperbolic spaces (Theorem 7.2.11 in Geometric group theory, Löh2017). That is, you can manage to find geodesics that are "close" to quasi-geodesics if the metric space is $\delta$-hyperbolic. Since $\mathbb{C}$ is not $\delta$-hyperbolic, one finds a counterexample.

A simple intermediate value argument will do: Prove that at one point the spiral is on one side of the ray, and at a later point the spiral is on the other side of the ray, and that the spiral cannot have gone around the end of the ray in between.

The following image will give you a general hint which points of the spiral to choose, though a thorough case analysis of the ray's location will be necessary: