Use geometry

But an even more elegant solution for this particular problem is as follows (sketched below; please fill in the details):

A rotation matrix is orthogonal, and therefore has determinant $\pm 1$; it is easy to see in this case that the determinant is 1. One of the eigenvalues must be 1 (it's a rotation in an odd-dimensional space); let the two others be $z_{1},z_{2}$. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have $1+z_{1}+z_{2}=1$ and $z_{1}z_{2}=1$, whence $\{z_{1},z_{2}\}=\{i,-i\}.$