In Polar Co-ordinates

\begin{aligned} \frac{\partial\psi^{\mathrm{R}}}{\partial x} &= C_0\frac{\sqrt{f}}{f’}\cdot \left( \cos\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot(\cos\phi\cos\theta+\sin\phi\sin\theta)\right) -\sin\theta\cdot\sin^{-1}\left(f’\cdot(-\cos\phi\sin\theta+\sin\phi\cos\theta)\right) \right) \\\
\frac{\partial\psi^{\mathrm{R}}}{\partial y} &= C_0\frac{\sqrt{f}}{f’}\cdot \left( \sin\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot(\cos\phi\cos\theta+\sin\phi\sin\theta)\right) +\cos\theta\cdot\sin^{-1}\left(f’\cdot(-\cos\phi\sin\theta+\sin\phi\cos\theta)\right) \right) \end{aligned}

Since these expression are constant in $R$, we can write \begin{aligned} \frac{\partial f}{\partial x} &= - \big(\frac{\sin\phi}{R}\big)\frac{\partial f}{\partial\phi} \\\
\frac{\partial f}{\partial y} &= \big(\frac{\cos\phi}{R}\big)\frac{\partial f}{\partial\phi} \end{aligned} where $f=\psi_x,\psi_y$.

Amplitudes for arbitrary orientation

The deflection is given as the vector $\nabla\psi^{\mathrm{R}}$, in a Cartesian co-ordinate system with axes aligned with the axes of the lens. We call this the lens frame. We are interested in the deflection vector $\vec{\alpha}(x,y)$ in a global frame, which shares the origin with the lens frame, but is rotated clockwise by an angle $\theta$. In other words, the lens is oriented at an angle $\theta$ (counterclockwise) in the global frame.

We will let $(x,y)$ denote the point in the global frame, and $(x’,y’)$ the same point in the lens frame. Hence \begin{aligned} \begin{bmatrix} x’ \\\\ y’ \end{bmatrix} &= \begin{bmatrix} \cos\theta & \sin\theta \\\
-\sin\theta & \cos\theta \end{bmatrix} \cdot \begin{bmatrix} x \\\\ y \end{bmatrix} \end{aligned} In other words, the $(x,y)$ coordinates are rotated clockwise. Similarly the deflection is given as $\nabla\psi^{\mathrm{R}}(x’,y’)$ in the lens frame, and $\vec{\alpha}(x,y)$ in the global frame. Thus, $\nabla\psi^{\mathrm{R}}$ has to be rotated counterclockwise, as \begin{aligned} \vec{\alpha}(x,y) &= \begin{bmatrix} \cos\theta & -\sin\theta \\\
\sin\theta & \cos\theta \end{bmatrix} \cdot \nabla\psi^{\mathrm{R}}(x’,y’) \end{aligned} This gives \begin{aligned} \vec{\alpha}(x,y) = C_0\frac{\sqrt{f}}{f’}\cdot \begin{bmatrix} \cos\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right) -\sin\theta\cdot\sin^{-1}\left(f’\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right) \\\
\sin\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right) +\cos\theta\cdot\sin^{-1}\left(f’\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right) \end{bmatrix} \end{aligned}