Let $S = \{z\in \mathbb{C} : |\arg(z)|\leq c\}$. This is a sector of the right half plane. We can make two immediate observations about points in this sector:
- If $z\in S$, then the real part of $z$ is nonnegative.
- There is a constant $A>0$ such that if $z = x + iy\in S$, then $|y|\leq Ax$. The constant $A$ is the absolute value of the slope of the bounding lines of the sector $S$.
Using the second bullet point, we see that if $z = x + iy\in S$, then $|z|^2 = x^2 + y^2 \leq (1+A^2)x^2,$ and hence $|z|\leq \sqrt{1+A^2}x = \sqrt{1+A^2} Re(z).$
We can then apply this to the $u_k$, to derive that $\sum_{k=1}^N|u_k|\leq \sqrt{1+A^2}\sum_{k=1}^N Re(u_k)\leq \sqrt{1+A^2} Re\sum_{k=1}^Nu_k\leq \sqrt{1 + A^2}\left|\sum_{k=1}^N u_k\right|.$ We know that $|\sum_{k=1}^N u_k|$ converges as $N\to \infty$, since $\sum u_k$ converges. Thus the right hand side is bounded. It follows that the sequence $\sum_{k=1}^N |u_k|$ is bounded in $N$. But this sum increases as $N\to \infty$, and all bounded increasing sequences converge. Thus $\sum |u_k|$ converges.