Question: I have trouble understanding the transition from the infinite integral of the Fourier transform
$$
\mathcal{F}f(v) = \int^\infty\_{-\infty}e^{ivk}f(k)dk
$$
to the discrete version 
$$
\mathbf{F}f\_n = \sum^{N-1}\_{k=0} f\_k e^{ink/N}.
$$
I first thought that one picks a finite subinterval of $\mathbb{R}$ and paritions it into $N$ pieces, and then somehow use periodicity of the integrand to extend this to infinity, but $e^{ivk}f(k)$ isn't necessarily periodic. This also doesn't make any sense if I want to calculate the transform for a single value $v$. So what does this finite sum actually represent?

Answer: 
Note that $\sqrt{\cos^2 \alpha - 1} = \sqrt{-(1-\cos^2 \alpha)} = \sqrt{- \sin^2\alpha} = i\sqrt{\sin^2\alpha}$...

So $$\lambda=\cos\alpha\pm\sqrt{\cos^2\alpha-1} = \cos \alpha \pm i\sin\alpha$$
$$\lambda=\cos\alpha\pm\sqrt{\cos^2\alpha-1}=\cos\alpha\pm\sqrt{(-1)(1-\cos^2\alpha)}=$$
$$=\cos\alpha\pm\sqrt{(-1)\sin^2\alpha}=\cos\alpha\pm\sqrt{(-1)}\cdot\sqrt{\sin^2\alpha}=$$
$$=\cos\alpha\pm i\sin\alpha$$