$\DeclareMathOperator{\trace}{trace}$ I am going through Rossmann's 'Lie Groups: An Introduction Through Linear Groups', and near the end of section 1.2 there is an example I don't quite understand (Example 9 in my edition). It is about solutions, $X$, to an equation of the form $$ \exp X = a$$. We assume $X$ is a $2\times 2$ matrix with trace zero. This allows us to obtain
$$ \exp X = (\cos \sqrt{\det{X}})1 + \frac{\sin \sqrt{\det{X}}}{\sqrt{\det{X}}}X.$$
So far so good. Now the book says that the above equation $$ \exp X = a$$ only has solutions when $\frac{1}{2} \trace a > -1$ or $a = -1$, and that in such a case for $-1 < \frac{1}{2} \trace a < 1$ the solution is given by
$$ X = \frac{\xi}{\sin \xi}(a-\frac{1}{2} \trace a 1),$$
where $\xi > 0$ satisfies $\cos \xi = \frac{1}{2} \trace a$. This is not clear to me. There seems to be some relationship here between $\trace a$ and $\det X$ which is not obvious to me. Could someone explain?
Edit: In light of one of the comment below, I'm expanding the post to include the follwing. @user1551 suggests that $\xi = \sqrt{\det{x}}$. Assuming $X$ to be in JNF form, we must have that the diagonal entries are such that
$$ \lambda_1 + \lambda_2 = 0 \rightarrow \lambda_2 = -\lambda_1$$
This would imply that
$$ \trace {a} = e^{\lambda_1} + e^{-\lambda_1}$$
and since $\det X = -\lambda_1^2$, we have
$$\trace{a} = 2 \cos\left({i \sqrt{\det{X}}}\right)$$.
But the equality does not hold without the $i$ in the argument. This is where I am confused.