\par\noindent{\bf Note. } If $r_{1,2}$ are a pair of complex conjugate numbers, then we are always in the ``otherwise'' case, and the two solutions are $y_1$ and its complex conjugate, $\overline{y_1}$. Note also that
\[ x^{\alpha\pm i\beta}=x^\alpha(\cos\log\beta\pm\sin\log\beta). \]
In that case it is more convenient to change to the ``real basis'' $(y_1+\overline{y_1})/2$ and $(y_1-\overline{y_1})/2i$ and get {\em real} solutions
\[