(Complex) angle between vectors in $\mathbb{C}^n$? For real vector spaces, the Cauchy-Schwarz inequality $|x\cdot y|\leq\Vert x\Vert \Vert y\Vert$ allows one to define a unique angle $\theta\in[0,\pi]$ between vectors $x$ and $y$ via $\cos^{-1}\frac{x\cdot y}{\Vert x\Vert \Vert y\Vert}$. But the Cauchy-Schwarz inequality is still valid for complex vectors. I was wondering if one can define the "angle" between complex vectors by the same inverse cosine formula. $x\cdot y=x^{\ast T} y$ is now complex, so the angle would become complex, but what is the "principal domain" of the angle to ensure single-valuedness of the inverse cosine function? Or does such a domain exist at all? I tried analyzing the formula
$$\cos^{-1}z=-i\log[z+i(1-z^2)^{1/2}]$$
without much success.
 A: The unmodified formula does result in a complex number, but it is hard to view it as an "angle." Two geometrically meaningful angles (although real, not complex) between complex vectors are the Euclidean angle $$\cos\theta_{E}\equiv\mathrm{Re}\left(\left\langle x,y\right\rangle \right)/\left(\left\Vert x\right\Vert \left\Vert y\right\Vert \right),$$ which is the angle between the vectors using the real inner product defined by the orthonormal basis in the decomplexified vector space, and the Hermitian angle $$\cos\theta_{H}\equiv\left|\left\langle x,y\right\rangle \right|/\left(\left\Vert x\right\Vert \left\Vert y\right\Vert \right),$$ which is the ratio of the orthogonal projection (using the complex inner product) of $x$ onto $y$ over the norm of $x$ (or the reverse), just as in the real case.
To answer the question about “principal domain,” the typical approach is to choose branch cuts in the complex plane of the domain of $\cos^{-1}$ to yield a continuous function, and then choose values continuous with a chosen real range. A common choice for $\cos^{-1}z$ is branch cuts at $(-\infty,-1)$ and $(1,\infty)$, with values chosen continuous with a real range of $[0,\pi]$. This results in a complex range equal to the strip of the complex plane with real part in $[0,\pi]$. The imaginary part of the range will also be limited on the edges: $\geq0$ for zero real part, and $\leq0$ for real part equal to $\pi$.
A: Yes, the regular definition of $$cos\ \theta = \frac{\overrightarrow{x}\cdot\overrightarrow{y^*}}{\|x\| \cdot \|y\|}$$ works for complex numbers, too. The principal domain = $[0, 2\pi]$.
