I have corrected the question in the following.
For $x_1$ and $x_2$ real vectors which span $V=\mathbb{R}x_1\oplus \mathbb{R} x_2$. we have a rotation $R$ on $V$ given by \begin{eqnarray*} R\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix} \end{eqnarray*} Now, consider $V^\mathbb C=\mathbb C\otimes_\mathbb{R}V=V\oplus iV$, it is $2$-dimensional $C$-vector space.
The question is:
How to choose a $\mathbb C$-basis such that $R$ acts on $V^\mathbb C$ as \begin{pmatrix} e^{i\theta}&\\ &e^{-i\theta} \end{pmatrix}?
The first column of $A$ must be a non-zero member of the kernel of the matrix $$ \pmatrix{ \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta } - e^{i\theta}\pmatrix{1&0\\0&1} = \pmatrix{ \frac{1}{2}\left(-e^{i\theta} + e^{-i\theta}\right) & \sin \theta\\ -\sin \theta & \frac{1}{2}\left(-e^{i\theta} + e^{-i\theta}\right) }\\ = \pmatrix{ i\sin \theta & \sin \theta\\ -\sin \theta & i\sin \theta } $$ Similarly, the second column of $A$ must be a non-zero member of the kernel of the matrix $$ \pmatrix{ \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta } - e^{-i\theta}\pmatrix{1&0\\0&1} = \pmatrix{ -i\sin \theta & \sin \theta\\ -\sin \theta & -i\sin \theta } $$