The contraction constant is k/4 for every k ≥ 2

For complex square contractions $A_1,\ldots,A_k$, with $|X|=(X^*X)^{1/2}$, the least constant $c_k$ such that $\left|\sum_{j=1}^kA_j\right|\le c_kI+\sum_{j=1}^k|A_j|$ holds in every dimension is $c_k=k/4$, for every integer $k\ge2$. For the upper bound, put $R=|\sum_jA_j|$ and $T=\sum_j|A_j|$; the vector Cauchy–Schwarz inequality gives $R^2\le k\sum_j|A_j|^2\le kT$, and then $T+\frac k4I-R\ge\frac1k\left(R-\frac k2I\right)^2\ge0$, without assuming the summands Hermitian. An explicit positive decomposition gives the same bound. Sharpness is attained by the rank-one contractions $A_j=e_1v_j^*$ in $\mathbb{C}^{2\times2}$ with $v_j=(1/2,(\sqrt3/2)\omega^j)^T$ and $\omega=e^{2\pi i/k}$: the vanishing sum of the $k$-th roots of unity leaves a gap of exactly $k/4$ at $e_1$. Hermitian $3\times3$ contractions attain it as well.