I've been reading about the Lawrence-Krammer representation of the braid group, which is a linear representation over the ring of bivariate Laurent polynomials $\mathbb{Z}[t, t^{-1}, q, q^{-1}]$. I've seen written in several places that with an appropriate choice of $t, q \in \mathbb{C}$, you can see that $\mathbb{Z}[t, t^{-1}, q, q^{-1}]$ is isomorphic to a subring of $\mathbb{C}$, and hence that the Lawrence-Krammer representation leads to a linear representation over the complex numbers, but no one seems to do this explicitly. How can you pick $t$ and $q$ so that the map $\mathbb{Z}[t, t^{-1}, q, q^{-1}] \to \mathbb{C}$ is injective?
I can see that if you had a univariate Laurent polynomial ring $\mathbb{Z}[t, t^{-1}]$, then you could pick $t$ to be any transcendental number: assume $t$ is transcendental, if $p(t) = q(t)$ for some distinct Laurent polynomials $p$ and $q$ of maximum degree $k$ then $t^k (p(t) - q(t)) = 0$ so $t$ must be algebraic, which is a contradiction.
If we were looking at a subset of bivariate Laurent polynomials which didn't have any 'cross-terms' then this would also be easy because we can map $t$ to a real transcendental number and $q$ to a purely imaginary transcendental number. However, for the general case I can't see how to make this work?
Your idea of picking real and purely imaginary transcendentals isn't necessary, because if you pick say, $e$ and $\pi i$, you still run into trouble because the latter squared is real: $(\pi i)^2 = - \pi^2$, Instead, what you need to know is that the transcendence degree of $\mathbb{C}$ as a field extension of $\mathbb{Q}$ is more than $2$, so that you can pick two algebraically independent transcendentals.
In fact, the transcendence degree of $\mathbb{C}$ is infinite (actually, it's $\mathfrak{c}$!), strictly for cardinality reasons. If you take a countable field such as $\mathbb{Q}$, and adjoin a transcendental element $\pi$, then $\mathbb{Q}(\pi)$ is actually still countable. To get the basic idea of why, consider which elements have been added: they are the polynomials in $\pi$, of which there are countably many in each degree. So by repeatedly taking these countable unions, and using the fact that every polynomial has finite degree, we see that $\mathbb{Q}(\pi)$ is a countable field. Now we can repeat the argument when we adjoint a second transcendental. By induction, any transcendental extension of finite transcendence degree is still countable.
But we know that $\mathbb{C}$ is uncountable, so there must be an infinite number of algebraically independent transcendentals over $\mathbb{Q}$. So your map is realized by picking two of them, and defining $1 \mapsto 1, t \mapsto \pi, q \mapsto \tau$, for two algebraically independent transcendentals $\pi, \tau$.
My guess is that the reason that nobody does this explicitly is because proving two transcendentals are algebraically independent is extremely difficult. I'm not an expert in this area, but I don't know of any two numbers proven to be algebraically independent over $\mathbb{Q}$.