I tried to determine the Lie algebra of $O(3, \mathbb C)$ but I think there is a mistake but I can't find it.
Here is my work:
Let $\mathfrak o$ denote the Lie algebra of $O(3, \mathbb C)$. The matrices in $O(3,\mathbb C)$ are those of determinant $\pm 1$. Hence $g \in \mathfrak o$ if and only if $e^{gt}$ has determinant $\pm 1$ if and only if $t \cdot \mathrm{trace}(g) $ is equal to zero or equal to $\pi i$ for all $t \in \mathbb R$ if and only if $\mathrm{trace}(g) = 0$.
That is, the case $t \cdot \mathrm{trace}(g)= i \pi$ is never satisfied because $t$ varies over $\mathbb R$. But then this would imply that the Lie algebra of $O(3, \mathbb C)$ equals the Lie algebra of $SL_3(\mathbb C)$ by the same argument.
Let $A:(-1,1)\to U(n)$ be a differentiable mapping such that $A(0) = I$. We know that for each $t \in (-1,1)$ that $A(t)A^\dagger(t) = I$. Take the derivative of both sides
\begin{equation*} \frac{d}{dt} A(t) A^\dagger(t) \;\; =\;\; \dot{A}(t)A^\dagger(t) + A(t)\dot{A}^\dagger(t) \;\; =\;\; 0. \end{equation*}
Now evaluate this at $t = 0$
$$ \left . \frac{d}{dt} A(t) A^\dagger(t) \right |_{t=0} \;\; =\;\; \dot{A}(0)A^\dagger(0) + A(0)\dot{A}^\dagger(0) \;\; =\;\; \dot{A}(0) + \dot{A}^\dagger(0) \;\; =\;\; 0 $$
where we use the fact that $A(0) = A^\dagger(0) = I$. This proves that $\dot{A}(0) = - \dot{A}^\dagger(0)$ so any tangent vector (or matrix rather) at the identity is a skew-Hermitian matrix. Therefore
$$ \mathfrak{u}(n) \;\; =\;\; \{A \in M_n(\mathbb{C}) \; | \; A = - A^\dagger\}. $$