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I am trying to compute the square root of the Y gate. $$Y_\theta =\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

There are many ways for doing this. I am keen on just doing the good old diagonalization and taking the square root of the eigenvalues. I compute the determinant of $Y-\lambda I$ and I set it to zero, and I get: $(cos\theta-\lambda)^2 + sin^2\theta=0$. This cannot have solutions, as a square cannot be negative.

**What am I doing wrong?**

5A square can be negative if it's a square of a complex number. Rotation gates are not self-adjoint, so they can have complex eigenvalues – Mariia Mykhailova – 2020-04-25T01:50:11.330