Show for a uniformly continuous $f:\mathbb{C} \mapsto \mathbb{C}$ that there exists $\alpha,\beta \in \mathbb{R}$ so that the following is true
$$\forall z\in\mathbb{C}: |f(z)| \le \alpha\cdot|z|+\beta$$
I have written down the definition of uniform continuity, which is
$$\forall \epsilon \gt 0\ \ \exists \delta\gt0 \ \forall z,w\in\mathbb{C} \ \text{with} \ |w-z|<\delta : |f(w) - f(z)|<\epsilon$$
But that is as far as I got.
Any tips are appreciated.
I will take $\epsilon=1$ in the definition of uniform continuity.
By squeezing the $\delta$ a bit, we can assume that it is the $\leq$-inequality instead of $<$ and $2\delta$ instead of $\delta$.
So let us write $|w-z|\leq 2\delta$ implying that $|f(w)-f(z)|\leq 1$.
For $|w-\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, then $|w|\leq|\delta e^{i\theta}|+\delta=2\delta$, so $|f(w)-f(0)|\leq 1$, so $|f(w)|\leq |f(0)|+1$.
For $|w-2\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, $|f(w)|\leq|f(2\delta e^{i\theta})|+1\leq |f(0)|+2$.
For $|w-n\delta e^{i\theta}|\leq\delta$, $\theta\in[0,2\pi]$, $|f(w)|\leq |f(n\delta e^{i\theta})|+1\leq\cdots\leq|f(0)|+n$.
But for any $z$, some $n$ and $\theta\in[0,2\pi]$ are such that $|z-n\delta e^{i\theta}|\leq\delta$, then $n\delta-|z|\leq\delta$ and hence $n\leq (|z|+\delta)/\delta$, and that $|f(z)|\leq|f(0)|+n\leq |f(0)|+(|z|+\delta)/\delta$.