The Claim
Let $L : \mathbb C\setminus\{0\} \to \mathbb C$ be a continuous function and define $f : \mathbb R \to \mathbb C, t \mapsto L(\mathrm e ^{\mathrm i t}) - \mathrm i t$. Show that $f \in 2\pi \mathrm i \mathbb Z$ for all $t \in \mathbb R$.
The beginnings of a proof
Let $\varepsilon > 0$ and $a = \mathrm i 2\pi m$, where $m \in \mathbb Z$. Then following from the continuity of $L$ and $-\mathrm i t$ (since $-\mathrm i t$ is a polynomial), as long as we choose $\delta > 0$ so that both $|L(t) - L(a)| < \varepsilon$ and $|-\mathrm i t + \mathrm i a| < \varepsilon$, we can write \begin{align} |f(t) - f(a)| &= |L(\mathrm e ^{\mathrm i t}) - \mathrm i t - ( L(\mathrm e ^ {\mathrm i (\mathrm i 2\pi m)}) - \mathrm i (\mathrm i 2 \pi m))|\\ &= |L(\mathrm e ^{\mathrm i t}) - L(\mathrm e ^ {- 2\pi m} ) - \mathrm i t \mathrm - 2 \pi m|\\ &\leq |L(\mathrm e ^{\mathrm i t}) - L(\mathrm e ^ {- 2\pi m} )| + | - \mathrm i t \mathrm - 2 \pi m|\\ &\quad\vdots\\ &< 2\varepsilon, \end{align} which would prove the claim. I'm just stuck at the three dots, and not even sure if I should be using the triangle inequality in this case.
Any hints?
An edit and a disclaimer
What I'm actually supposed to be proving is that there is no such continuous function $L$ for which the equation $\mathrm e^{L(z)}=z$ holds.
I'm being told I should assume the opposite, show that the claim in this post holds and deduce, that as a result $f$ would have to be constant using Bolzano's theorem. This should result in a contradiction because of the fact, that $f(t+2π)=f(t)$ for all $t\in \mathbb R$.
Edit number 2
I've just realized that it couldn't possibly be that $a \in \mathrm i 2 \pi \mathbb Z$ (at least not if I want the $\varepsilon$-$\delta$ proof above to work), since $f: \mathbb R \to \mathbb C$. I wonder if our lecturer made a typo in ths case.