An element of $SL(2,\mathbb{R})$ An element $A$ of $SL(2,\mathbb{R})$ is called an elliptic element if $|\text{tr}(A)|<2$.

Find the relationship between an elliptic element of $SL(2,\mathbb{R})$ and rotation.

As $|\text{tr}(A)|<2$ the characteristic equation of $A$ does not have real roots, so it has no real eigenvalue. But I am unable to go ahead.
 A: Let $g$ be your elliptic element. Since the eigenvalues are complex conjugate, $g$ will have an eigenvector $\binom z1\in\Bbb C^2$ with the imaginary part of $z$ positive.
It can be shown that you can always find some $h\in{\rm SL}_2(\Bbb R)$ such that $h\binom i1=c\binom z1$ for some $c\in\Bbb R$.
Thus $h^{-1}gh$ will have $\binom i1$ as eigenvector.
The final step is to show that the elements of ${\rm SL}_2(\Bbb R)$ having $\binom i1$ as an eigenvector are precisely the elements in the subgroup ${\rm SO}_2(\Bbb R)$, i.e. the rotations.
I leave the details as an exercise. 
A: Here is a second proof. Let $\theta\in\Bbb R\setminus\pi\Bbb Z$ be such that $\mathrm{Tr}(A)=2\cos(\theta)$. The characteristic polynomial of $A$ equals $\chi_A(X)=X^2-\mathrm{Tr}(A)X+\det(A)=(X-e^{i\theta})(X-e^{-i\theta})=\chi_{R_{\theta}}(X)$ where
$$
R_{\theta}=\begin{pmatrix}
\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)
\end{pmatrix}
$$
Since $\theta\notin\pi\Bbb Z$, $e^{i\theta}\neq e^{-i\theta}$ and $A$ and $R_{\theta}$ are diagonalzable (over $\Bbb C$) and conjugate (as complex matrices) to
$$\begin{pmatrix}
e^{i\theta}&0\\0&e^{-i\theta}
\end{pmatrix}$$
We conclude by invoking the general fact that two real matrices that are conjugate as complex matrices are conjugate as real matrices.
