The smooth map
$$f_a :\mathbb R \to \mathbb C, f_a(t) = ae^{(1+i)t} = ae^te^{it}$$
is an injective immersion:
We have $f_a'(t) = (1+i)ae^{it} = (1+i)f_a(t) \ne 0$, thus $f_a$ is an immersion.
$f_a(t) = f_a(s)$ implies $ae^t = \lvert f_a(t) \rvert = \lvert f_a(s) \rvert = ae^s$, thus $t = s$. Hence $f_a$ is injective.
$f_a$ embeds $\mathbb R$ as a smooth submanifold of $\mathbb C = \mathbb R^2$. Since $f_a(\mathbb R) = \Gamma_a$, the set $\Gamma_a$ is a smooth submanifold of $\mathbb C$. Note that $\Gamma_a$ is an infinite spiral around the origin.
$f_a(\mathbb R) = \Gamma_a$ is obvious. To show $f_a$ is an embedding, it suffices to show that $f_a : \mathbb R \to \Gamma_a$ is a homeomorphism (recall 1.). Clearly $\lim_{t \to \infty} f_a(t) = \infty$ and $\lim_{t \to -\infty} f_a(t) = 0$. Thus $f_a$ has a continuous extension $\bar f_a : [-\infty,\infty] \to \hat{\mathbb C}$. Here $[-\infty,\infty]$ is the extended real line (which is homeomorphic to $[-1,1]$) and $\hat{\mathbb C} = \mathbb C \cup \{\infty\}$ is the Riemanns sphere. Since $\bar f_a$ is injective with compact domain, it is a topological embedding and so is its restriction $f_a$.
$\Gamma_n \cap \Gamma_m = \emptyset$ for all integers $n, m$ with $n \ne m$.
This is highly non-trivial. To prove it, we have to show that
$$ne^te^{it} = me^se^{is} \tag{1}$$
does not have a solution for $n \ne m$. $(1)$ implies that $e^{i(t-s)}$ must be a positive real number which is possible only if $e^{i(t-s)} = 1$, i.e. $s = t + 2k\pi$ for some $k \in \mathbb Z$. This gives
$$me^t = ne^{t +2k\pi}$$
and thus
$$m/n = e^{2k\pi} = (e^\pi)^{2k}. \tag{2}$$
It is known that $e^\pi$ is transcendental, thus the RHS of $(2)$ is transcendental for $k \ne 0$. Thus $(2)$ is possible only for $k=0$, i.e. for $n = m$.
The infinite union $\Gamma = \Gamma_1 \cup \Gamma_2 \cup \ldots$ is the disjoint union of pairwise disjoint smooth submanifolds of $\mathbb C$.
Consider the space $M = \mathbb R \times \mathbb N$. This is a (non-connected) smooth $1$-manifold. The map
$$f : M \to \mathbb C, f(t,n) = f_n(t)$$
is an injective immersion. We shall prove that $f$ is not a topological embedding (which means that $\Gamma$ is not a smooth submanifold of $\mathbb C$).
Let $d_n = \ln (n+1) - \ln n = \ln \frac {n+1}{n}$. Then $d_n \to 0$ as $n \to \infty$. Since $(\ln n)$ is strictly increasing with $\ln n \to \infty$ as $n \to \infty$, we see that for each $k \in \mathbb N$ there exists a unique $n_k \in \mathbb N$ such that $\ln (n_k +1) > 2\pi k \ge \ln n_k$. The sequence $(n_k)$ is strictly increasing with $n_1 > 1$ and we have $\ln n_k - 2\pi k \to 0$ as $k \to \infty$. The set
$$C = \bigcup_{k=1}^\infty \{(-2\pi k,n_k)\}$$
is a closed subspace of $M$. If $f$ were a topological embedding, then $f(C)$ should be a closed subspace of $\Gamma$. But
$$f(-2\pi k,n_k) = n_ke^{-2\pi r_{n_k}} = e^{\ln n_k - 2\pi r_{n_k}} \to e^0 = 1 = f(0,1) \in \Gamma_1. \tag{3}$$
Since $f(C)$ does not intersect $\Gamma_1$, $(3)$ shows that $f(C)$ is not closed in $\Gamma$.
Therefore $\Gamma$ is not a smooth submanifold of $\mathbb C$.
Remark.
In Step 3 we used the fact that $e^\pi$ is transcendental to show that $\Gamma_n \cap \Gamma_m = \emptyset$ for $n \ne m$. This gives a better understanding of $\Gamma$. But for the proof that $\Gamma$ is not a smooth submanifold of $\mathbb C$ we do not need to know this. We can argue as follows: Either the $\Gamma_n$ are pairwise disjoint (in which case the above argument works) or $\Gamma_n \cap \Gamma_m \ne \emptyset$ for some numbers $n \ne m$. In that case we have a (unique) intersection point which obviously prevents $\Gamma$ from being a $1$-dimensional manifold.
