Relation between transcendental numbers and computational complexity?
One other way to look at this, which brings in potentially all complexity classes above $\mathsf{E} = \mathsf{DTIME}(2^{O(n)})$, is to consider real numbers in their binary expansion. Any real number whose binary expansion doesn't end with $0^\infty$ or $1^\infty$ - i.e., which is not a dyadic rational - has a unique binary expansion. We can treat this binary expansion as an element of $\{0,1\}^{\mathbb{N}}$, which in turn we can view as a subset of $\mathbb{N}$, and therefore (under the standard identification of $\mathbb{N}$ with $\Sigma^*$ for $\Sigma$ a finite alphabet) a language. One can then ask about the complexity of this language. (Dyadic rationals correspond to languages which are only finitely different from $\emptyset$ or from $\Sigma^*$, so aren't all that interesting anyways...)

Note that if a language $L$ has its characteristic function $\chi_L$ being real-time computable, then $L$ itself is in $\mathsf{E}$: given input $x$ of length $n$, suppose that $x$ is the $N$-th string in length-lexicographic order (that is, the string $x \in \Sigma^*$ corresponds to the integer $N \in \mathbb{N}$ under the standard identification). To decide whether $x \in L$, use the real-time machine to output the first $N$ bits of $\chi_L$ and read off the $N$-th bit. This takes $O(N)$ time; but $|x| = \log_{|\Sigma|} N$, so this is $O(|\Sigma|^{|x|}) = 2^{O(|x|)}$ time. 

(As far as I can tell the converse need not hold. Or at least, the naive proof that I was thinking of doesn't work. That is, suppose $L \in \mathsf{E}$. The naive way  decide the first $N$ bits of $\chi_L$ takes $\approx N \cdot 2^{O(\log N)} = N^{1 + c}$ time for $c &gt; 0$.)

Using this identification, you can in principle use all complexity classes that aren't contained in $\mathsf{E}$ to try to understand the "complexity" of any non-real-time-computable transcendental numbers. However, I don't know how to relate complexity properties of the language $L$ to whether $\chi_L$ is algebraic or transcendental. Strikes me as a very interesting question. As pointed out by Emil Jerabek in the comments, using this identification, all algebraic numbers are in the linear-time version of the counting hierarchy $\mathsf{CH}$, hence also in $\mathsf{E}$. So this approach seems mainly useful for distinguishing the complexity of transcendental numbers from one another, rather than for distinguishing e.g. the complexity of real-time computable and algebraic numbers.