Is the following proposition true?
Let $f_n \colon U\subset\mathbb{C}\longrightarrow \mathbb{C}$ be a sequence of continuous (or holomorphic) functions such that:
$\bullet$ $\sum f_n$ is absolutely convergent
$\bullet$ $\sum f_n$ is uniformly convergent
Then $\sum |f_n|$ is uniformly convergent.
Any help would be appreciated.
The absolute convergence of $\sum f_n$ and the uniform convergence of $\sum f_n$ do not imply the uniform convergence of $\sum \lvert f_n\rvert$, whether the $f_n$ are holomorphic or not.
Let $a_n = \frac{(-1)^n}{n}$ for $n\in\mathbb{N}\setminus \{0\}$, and $f_n(z) = \frac{a_n}{n^z}$, where for the exponential $n^z$ we use the real value of $\log n$. Then we have
$$\lvert f_n(z)\rvert = n^{-(1+\operatorname{Re} z)},$$
so the series $\sum_{n=1}^\infty f_n(z)$ is absolutely convergent on the right half-plane, and the convergence is absolute and uniform on the closed half-plane $H_t := \{ z : \operatorname{Re} z \geqslant t\}$ for every $t > 0$.
For $0 < \alpha < \pi/2$, let $U_\alpha = \{z\in \mathbb{C}\setminus\{0\} : \lvert \operatorname{Arg} z\rvert < \alpha\}$. Then $\sum_{n=1}^\infty f_n(z)$ converges uniformly on $U_\alpha$, but $\sum_{n=1}^\infty \lvert f_n(z)\rvert$ does not converge uniformly on $U_\alpha$.
To see the latter assertion, note that
$$\lim_{z\to 0} \sum_{n=1}^N \lvert f_n(z)\rvert = \sum_{n=1}^N \frac{1}{n} > \log N,$$
so the sequence of partial sums is not uniformly bounded on $U_\alpha$, but since every $\lvert f_n(z)\rvert$ is bounded (by $1/n$), if the series were uniformly convergent, the sequence of partial sums would be uniformly bounded.
To see that $\sum_{n=1}^\infty f_n(z)$ is uniformly convergent on $U_\alpha$, let $R_k = \sum_{n=k}^\infty a_n$. Since $(a_n)$ is alternating with monotonically decreasing modulus, we have $\lvert R_k\rvert \leqslant \lvert a_k\rvert = \frac{1}{k}\to 0$. Then we compute, writing $s = \operatorname{Re} z$,
\begin{align} \left\lvert \sum_{n=K}^M \frac{a_n}{n^z}\right\rvert &= \left\lvert\sum_{n=K}^M \frac{R_n - R_{n+1}}{n^z} \right\rvert\\ &= \left\lvert \frac{R_K}{K^z} - \frac{R_{M+1}}{M^z} - \sum_{n=K+1}^M R_n\left(\frac{1}{(n-1)^z} - \frac{1}{n^z}\right)\right\rvert\\ &\leqslant \frac{\lvert R_K\rvert}{K^s} + \frac{\lvert R_{M+1}\rvert}{M^s} + \sum_{n=K+1}^M \lvert R_n\rvert\left\lvert \frac{1}{(n-1)^z} - \frac{1}{n^z}\right\rvert\\ &\leqslant \frac{1}{K} + \frac{1}{M} + \frac{1}{K}\sum_{n=K+1}^M \left\lvert \frac{1}{(n-1)^z} - \frac{1}{n^z}\right\rvert \tag{$\ast$}\\ &\leqslant \frac{1}{K} + \frac{1}{M} + \frac{\lvert z\rvert/s}{K}\left(\frac{1}{K^s} - \frac{1}{M^s}\right)\\ &\leqslant \frac{1}{K}\left(2+\frac{\lvert z\rvert}{s}\right)\\ &\leqslant \frac{1}{K}\left(2 + \frac{1}{\cos\alpha}\right), \end{align}
which shows the uniform convergence on $U_\alpha$.
In $(\ast)$, we used
$$\frac{1}{(n-1)^z} - \frac{1}{n^z} = z\int_{n-1}^n \frac{dt}{t^{z+1}}$$
from which we obtain the estimate
$$\left\lvert \frac{1}{(n-1)^z} - \frac{1}{n^z}\right\rvert \leqslant \lvert z\rvert\int_{n-1}^n \frac{dt}{\lvert t^{z+1}\rvert} = \lvert z\rvert\int_{n-1}^n \frac{dt}{t^{s+1}} = \frac{\lvert z\rvert}{s}\left(\frac{1}{(n-1)^s}-\frac{1}{n^s}\right),$$
and that telescopes. Finally, $K^{-s} - M^{-s} < K^{-s}\leqslant 1$.