For the inner product space $M_{2\times2} (\Bbb{C})$, with inner product $\langle A,B\rangle = \operatorname{tr}(AB^*)$, find the angle between
$$A=\begin{bmatrix}2+i&1\\-1&0\end{bmatrix} $$
$$B=\begin{bmatrix}3-2i&1+i\\0&-1\end{bmatrix}$$
I tried to use $$\cos(z)=\frac{\langle A,B\rangle}{\|A\|\|B\|}=\frac{5+6i}{4\sqrt{7}}$$ but how can I find the angle? is it supposed to be a real number?
The inverse of the cos function is defined for complex arguments as well.
$$\operatorname{arccos}(z)=-i\operatorname{Log}\left(iz+\sqrt{1-z^2}\right).$$
Where Log is the so called complex logarithm defined as
$$\operatorname{Log}(u)=\ln(\mid u\mid)+i\operatorname{atan2}(\Im(u),\Re(u)).$$
And finally $\operatorname{atan2}$ is defined as
So the complex angle in the OP's case would be a complex number:
$$\operatorname{arccos}\left(\frac{5+6i}{4\sqrt{7}}\right).$$