I am *not* so satisfied with this the following
*definition*.^{1}

I remembered the proof for convergence of

for real numbers. I *didn’t* know if this can be extended to complex
numbers. Therefore, I thought about the absolute convergence of
complex-valued series. It’s expected that many proofs are similar to
their real counterparts, such as the result that absolute convergence
implies convergence. In real numbers, this result makes use of the
Triangle Inequalty and Cauchy Convergence Criterion, and the key step
is

Since the proof of the above statement for real numbers requires
Bolzano–Weierstrass Theorem, which is about the sequential
compactness of sequences of real numbers, I was *stuck* at this point.

Finally, I read another book, which said that if $(z_n)$ is a Cauchy sequence, and $\forall n \in \N$, $u_n := \Re(z_n)$ and $v_n := \Im(z_n)$, $\forall \varepsilon > 0, \exists N \in \N$ such that $\forall m,n \le N$,

Then $(u_n)$ and $(v_n)$ are real-valued Cauchy sequences, which are
convergent.^{2} This guarantees the convergence of $(z_n)$ in
the complex plane.

To establish the absolute convergence of $\exp z$, we need the root test. The proofs can be borrowed from their counterparts in the set of real numbers. Ahlfors leaves the proof for $\sqrt[n]{n!} \to \infty$ to readers. I find Dan’s proof pretty easy.