Absolutely convergent series of complex functions.

I have to do the following excercise:

Let $$\{f_n(z)\}_{n\in\mathbb{N}}$$ a sequence of complex functions, and let $$\sum_{n=1}^\infty f_n(z)$$.

Prove that: if $$\sum_{n=1}^\infty |f_n(z)|$$ converges, then $$\sum_{n=1}^\infty f_n(z)$$ converges.

I know how to prove it for a series $$\sum_{n=1}^\infty z_n$$ of complex numbers with $$z_n=x_n+iy_n$$ because if $$\sum_{n=1}^\infty |z_n|$$ converges, one can observe that $$|x_n|<|z_n|$$ and $$|y_n|<|z_n|$$ then by the comparison criteria the real numbers series $$\sum_{n=1}^\infty |x_n|$$ and $$\sum_{n=1}^\infty |y_n|$$ converge and we know for real series that this implies that $$\sum_{n=1}^\infty x_n$$ and $$\sum_{n=1}^\infty y_n$$ converge.

If we call $$R_n=\sum_{k=1}^n x_n$$, $$I_n=\sum_{k=1}^n y_n$$ and $$S_n=\sum_{k=1}^n z_n$$.

And $$\lim_{n \rightarrow \infty}R_n=x$$, $$\lim_{n \rightarrow \infty}I_n=y$$, then

$$\lim_{n \rightarrow \infty}S_n=\lim_{n \rightarrow \infty}R_n+i\lim_{n \rightarrow \infty}I_n=x+iy.$$

Then $$S_n$$ converges and $$\sum_{n=1}^\infty z_n$$ does as well.

Is it enough to call $$\{w_n\}=\{f_n(z)\}$$ in my original problem and just apply this proof?

Yes, that would be correct. On the other hand, you don't have to decompose your series into real and imaginary part. Suppose that $$\sum_{n=1}^\infty\lvert z_n\rvert$$ converges. Take $$\varepsilon>0$$. Then there is a natural $$N$$ such than$$m\geqslant n\geqslant N\implies \sum_{k=n}^m\lvert z_k\rvert<\varepsilon,$$and therefore, by the triangle inequality,$$m\geqslant n\geqslant N\implies\left\lvert\sum_{k=n}^mz_k\right\rvert<\varepsilon.$$Therefore, by Cauchy's criterion, the series $$\sum_{n=1}^\infty z_n$$ converges too.