Start with $\mathbb{R}^n$. If I take a vector $v \in \mathbb{R}^n$ of norm $1$, it is not difficult to construct an explicit basis of $\mathbb{R}^n$ of the form $B_v = (v, v_2, \cdots, v_n) \in SO(n)$. If I consider the map $B : v \in \mathbb{S}^{n-1} \to B_v \in SO(n)$, then the map $B$ cannot be continuous on $\mathbb{S}^{n-1}$, put can be continuous on $\mathbb{S}^{n-1}$ minus a north pole. An explicit construction is to take the $n-1$ partial derivatives of the stereographic projection (that gives a basis of the tangent plane of $\mathbb{S}^{n-1}$ at $v$).
My question is: what about $\mathbb{C}^n$? Is there a "canonical (and smooth) way" to complete a vector $z \in \mathbb{C}^n$ with $\| z \| = 1$ into a basis of $\mathbb{C}^n$?
Thanks
First of all, your statement is not correct.
For some $n$, namely $n=2,4,8$ such a map exists : it is given by a (orthonormal) trivialisation of the tangent bundle of the sphere $S_{n-1}$.
For instance in dimension 2, it is given by $\nu(u)=iu$ where $i^2=-1$ after identifying $\bf R^2$ with $\bf C$. In dimension 4, after identifiying $\bf R^4$ with the quaternions the matrix of the right multiplication by $q$ is orthogonal $q,Iq,Jq,Kq$. This is a deep theorem that the tangent bundle of the sphere $S_{n-1}$ is trivial only in dimension $2,4,8$ (for 8 use octonions).(Bott-Milnor-Kervaire)
Now your question reduces to understand if a certain complex bundle over $S^{2n-1}$ is trivial or not. At a point $p$ on the sphere, this bundle is the vectors orthogonal to $p$ in the hermitian sense, and the triviality implies also the triviality of the tangent bundle of the sphere (add the direction $ip$).
In other words, the existence of such a map implies $2n-1= 1,3,7$, and $n=1,2,4$, and one can construct the map $\nu$ with the same argument (use quaternions, octonions).