The group $SU(n, \mathbb{C})$ is the set of $n \times n$ complex matrices $Q$ such that $\det Q = 1$ and $Q\overline{Q}^{T} = 1$. The Lie algebra is the set of traceless anti-Hermitian matrices.
To compute the Lie algebra, we have: Write $Q = I + \epsilon K$ be a matrix in $SU(n, \mathbb{C})$. Then as $Q\overline{Q}^{T} = I$, $I + \epsilon(\overline{K}^{T} + K) + \epsilon^{2}K\overline{K}^{T} = I$. Letting $\epsilon \rightarrow 0$, we have $\overline{K}^{T} + K = 0$ and $K$ must be anti-Hermitian. How do I find that the matrices in the Lie algebra must also be traceless?