While solving the equation \begin{equation} x^2 + a = 0 \end{equation} in $\mathbb{C}$, where $a$ is a complex number, i wanted to explicitate the geometrical intuition behind this and overthought the whole thing as follows.
Let $z_1$ and $z_2$ be the solutions to the equation: we know that $z_1+z_2 = 0$ and $z_1z_2 = a$. So $z_1$ and $z_2$ are opposite points wrt the origin, and it has to be possible to get to them by rotating and scaling the points $(1,0)$ and $(-1,0)$, i.e. we can surely find a real number $\lambda$ and a rotation matrix $R_\theta$ such that \begin{equation} \lambda R_\theta\cdot\left(\begin{matrix}1\\0\end{matrix}\right) = z_1\qquad\text{and}\qquad \lambda R_\theta\cdot\left(\begin{matrix}-1\\0\end{matrix}\right) = z_2 \end{equation} All seems nice and tame until I multiply the two equations: \begin{equation} \lambda \left( \begin{matrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{matrix} \right) \left(\begin{matrix}1\\0\end{matrix}\right) \cdot \lambda \left( \begin{matrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{matrix} \right) \left(\begin{matrix}-1\\0\end{matrix}\right) = z_1z_2 = a \end{equation} because I get \begin{equation} \lambda \left( \begin{matrix} \cos\theta\\ \sin\theta \end{matrix} \right) \cdot \lambda \left( \begin{matrix} -\cos\theta\\ -\sin\theta \end{matrix} \right) = -\lambda^2(\cos^2\theta+\sin^2\theta)=-\lambda^2=a \end{equation} which of course is true, but this equation tells us that $a$ is a negative real. It seems to me that this calculation hides the angle information in $\lambda$: indeed, if we allow $\lambda$ to be complex then of course we get the two solutions simply by solving $\lambda^2 = -a$.
At which step did the information on the angle get erased? If I want to keep $\lambda$ real as a scaling factor, how can I retrieve the information on $\theta$?