Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$ I think that this is an exercise. I can not find a solution.
We can define Lie bracket multiplication on $\mathbb{C}^3$ : $$ x\wedge y $$ where $x=(x_1,x_2, x_3)$, $y=
(y_1,y_2,y_3)$, and $\wedge $ is the wedge product we know.
Consider the Lie algebra $\mathfrak {sl}(2,\mathbb{C})= \{ X\in M_2(\mathbb{C}) \mid\ {\rm Trace} (X) =0\}$
and $$
e= \left(
  \begin{array}{cc}
    0 & 1 \\
    0 & 0 \\
  \end{array}
\right),\ f= \left(
  \begin{array}{cc}
    0 & 0 \\
    1 & 0 \\
  \end{array}
\right),\ h= \left(
  \begin{array}{cc}
    1 & 0 \\
    0 & -1 \\
  \end{array}
\right). $$ Note that $$ [e,f]=h,\ [e,h]=-2e,\
[f,h]=2f.$$
Here, the problem is to find an explicit isomorphism between $\mathfrak {sl}(2,\mathbb{ C})$ and $\mathbb{C}^3$.
Thank you in advance.

To Learner,
I am trying to find a isomorphism from ${\bf R}^3$ to $sl(2,{\bf
  R})$. But from the comments of Avitus, I realized that I should
  find a suitable basis in ${\bf C}^3$.
Note that $\{ e,f,h\} $ of $sl(2,{\bf R})$ in original post is a
  basis in $sl(2,{\bf C})$ : And these elements satisfy the relation
  $$\ast\ [e,f]=h,\ [e,h]=-2e,\ [f,h]=2f $$
So remaining thing is to make a basis in ${\bf C}^3$ : I do not
  know who will find through $reasonable\ way$. My way is try and
  error, i.e., just trying.
(1) If $v_k=\sum_{i=1}^3 a^k_i e_i,\ a_i^k\in {\bf C}$ where $e_i$ is a canonical basis in ${\bf R}^3$,
   then let $\phi
  (v_1)=e,\ \phi (v_2)=f,\ \phi (v_3)=h$ so that we have an equation
  for $a^k_i$. Then we solve it (I believe that this way would work).
(2) Another way : Note that from $\ast$ we have $$ [e+f,h]=-2(e-f) $$
 And $$ [e+f,e-f]=-2h $$
That is we have $v=e+f,\ w=h,\ x=e-f $ s.t. they are basis for $sl(2, {\bf R})$ and $$
    [v,w]=-2x,\ [w,x]=2v,\ [x,v]=2w $$
This relation is similar to $(\{ e_i\},\wedge )$ so that we let
   $$
   \phi(v)=ae_1,\ \phi (w)=be_2,\ \phi (x)=ce_3 $$ So from calculation we have
    $$\frac{1}{2} (a,b,c)=( \pm i,\pm i,1)$$
