Can $g(t) = \infty$ in $\hat{\mathbb{C}}$?

Question

Let $z\in\mathbb{C}$ be a constant such that $\Re z>0$ (its real part is positive) and $\Im z \neq 0$ (its imaginary part is not zero), then let $f$ and $g$ be functions such that:

$$ \begin{aligned} f\colon \, \mathbb{R}\to\mathbb{C} \\ f(t)= e^{zt} \end{aligned} \quad\quad\quad\quad\quad \begin{aligned} g\colon \, \mathbb{R}\to\hat{\mathbb{C}} \\ g(t)= e^{zt} \end{aligned} $$

where $\hat{\mathbb{C}}$ is the extended complex plane ($\hat{\mathbb{C}} = \mathbb{C}\,\cup\,\left\lbrace \infty \right\rbrace$).

I can easily prove that:

$$ \lim_{t\to\infty} |f(t)| = \lim_{t\to\infty} |e^{zt}| = \lim_{t\to\infty} e^{t\Re z} \to \infty \quad\Longrightarrow\quad \not\exists \lim_{t\to\infty} f(t)$$

But when I want to do the same for $g(t)$:

$$ \lim_{t\to\infty} |g(t)| = \lim_{t\to\infty} |e^{zt}| = \lim_{t\to\infty} e^{t\Re z} \overset{??}{=} \infty \quad\Longrightarrow\quad \exists \lim_{t\to\infty} g(t) = \infty ?$$

So, I have two existential doubts:

1. Can a limit (and an expression in general) be equated to infinity, rigorously, when working with $\hat{\mathbb{C}}$? This would be to say that $ z=\infty$ is valid in $\hat{\mathbb{C}}$, rigorously. Therefore, it is rigorously valid to say that (notice the equal symbol):

$$\lim_{t\to\infty} |g(t)| = \infty$$

2. Does the expression $\lim_{t\to\infty} g(t) = \infty$ is valid? I ask because, although $\lim_{t\to\infty} |g(t)| = \infty$, it happens that $$\lim_{t\to\infty} e^{i t\Im z} = \lim_{t\to\infty} \cos(t\Im z)+i\sin(t\Im z) $$ is oscillating between all the complex signs. So, I conclude that $$\lim_{t\to\infty} g(t) = \lim_{t\to\infty} |g(t)|\left( \cos(t\Im z)+i\sin(t\Im z) \right) $$ remains oscillating in all possible infinites (in all directed infinites) and therefore it is rigorously valid to say that $$\lim_{t\to\infty} g(t) = \infty$$.

Answer 1

For the sake of completeness, let me post the essence of my comments above as an answer. To talk about limits rigorously, one must clearly understand the underlying topology. The extended complex plane is usually topologized as the one-point compactification of the complex plane. This means that a neighborhood of the point $\infty$ is the complement of a compact (closed and bounded) set in the complex plane. With this in mind, it is easy to see that a function $f:\mathbb R \to \hat {\mathbb C}$ satisfies $f(t) \to \infty$ as $t \to \infty$ iff $|f(t)| \to \infty$, which answers your question.