The task is to evalute $$ \int\limits_{\mathbb{R}^2} \frac{e^{i \langle \xi, x \rangle} d\xi}{ \langle\xi,\theta\rangle}, \;\;\; \theta \in \mathbb{C}^2 \setminus ( \mathbb{S}^1 \cup \left\{ 0 \right\} ), \;\;\; \theta_1^2 + \theta_2^2 = 1,\;\;\; x\in \mathbb{R}^2 $$ Obviously it is the Fourrier transform of $f(\xi) = \langle \xi,\theta \rangle ^ {-1}$. I tried to make the change $y_{1} = \langle\xi,\Re \theta\rangle, \;\; y_2 = \langle \xi, \Im \theta \rangle$, but i faced with difficult calculations.
Update: let $$ f(x) = \frac{1}{2\pi i}\int\limits_{\mathbb{R}^2} \frac{e^{i \langle \xi, x \rangle} d\xi}{ \langle\xi,\theta\rangle} $$ Then, using inverse Fourrier transform we get $$ i\langle\theta,\xi\rangle \hat{f}(\xi) = \frac{1}{\sqrt{2\pi}} $$ Taking Fourier transform now we recieve $$ \langle \theta, \nabla f(x) \rangle = \delta(x) $$ In other words, f(x) is a Green function for operator $L = \langle \theta, \nabla \rangle$.
The following calculation uses the standard assumptions of the extension of the Fourier transform to the set of tempered distributions and the use of the Cauchy principal value in the definition of $\frac{1}{\xi}$. Since the Lebesgue measure of $\mathbb{R}^2$ is invariant under plane rotations, the integral is invariant under a mutual rotation of the vectors $x$ and$\theta$. Applying the rotation matrix:
$ \left (\begin{array}{cc} \theta_1 &\ \theta_2\\ -\theta_2 &\ \theta_1 \end{array} \right )$,
(note that this is an orthogonal matrix since $\theta_1^2+\theta_2^2=1$) to both $x$ and $\theta$, transforms $\theta$ to $(1,0)$ and $x$ to $(\theta_1 x_1 +\theta_2 x_2 ,-\theta_2 x_1 +\theta_2 x_2)$.Thus the integral becomes:
$\int_{\mathbb{R}^2} \frac{e^{i(\theta_1 x_1 +\theta_2 x_2)\xi_1+i( -\theta_2 x_1 +\theta_2 x_2)\xi_2}}{\xi_1} d\xi_1 d\xi_2= -i \pi sgn(\theta_1 x_1 +\theta_2 x_2) \delta( -\theta_2 x_1 +\theta_2 x_2)$
Where the Fourier transform table (for distributions) from the Wikipedia article was used.