Homotopy, Stokes Theorem and Orientation

Question

I have a problem in which the theory and the computation disagree about a minus sign. My question requires a little setting up. I have a complex valued 2-form $$ \omega = \alpha(\xi_1,\xi_2)\, \mathrm{d}\xi_1 \wedge \mathrm{d}\xi_2 $$ such that $\alpha=\alpha(\xi)$ is an entire function of $\xi=(\xi_1,\xi_2)\in \mathbf{C}^2$. Consequently $\mathrm{d}\omega=0$ since all the $\bar{\partial}$-derivatives of $\alpha$ vanish. Let $U\in \mathrm{SU}(2)$ and define $f$ by $$ f:\mathbf{C}^2\rightarrow \mathbf{C}^2: \xi \mapsto e^{\mathrm{i}\pi/2} U\xi. $$ Let $h:[0,1]\times \mathbf{C}^2\rightarrow \mathbf{C}^2$ be a homotopy between $f$ and the identity, so $h(0,\xi)=\xi$ and $h(1,\xi)=f(\xi)$ and let $B_R$ be the ball of radius $R$ centered about the origin in $\mathbf{R}^2\subset\mathbf{C}^2$. Since $\omega$ is closed and $\mathrm{d}(h^*\omega) = h^*(\mathrm{d}\omega)$ we have by Stokes theorem $$ \begin{aligned} 0 \quad & = \int_{B_R\times [0,1]} \mathrm{d}(h^*\omega) \\ & = \int_{\partial B_R\times[0,1]} h^*\omega + \int_{B_R\times\{1\}} h^*\omega - \int_{B_R\times\{0\}} h^*\omega \\ & = \int_{\partial B_R\times[0,1]}h^*\omega + \int_{B_R} f^*\omega - \int_{B_R} \omega. \end{aligned} $$ I have shown for my $f,h$ and $\alpha$ the first integral vanishes in the limit as $R\rightarrow \infty$. So we get $$ \int_{\mathbf{R}^2} f^*\omega = \int_{\mathbf{R}^2} \omega. $$ Computing the pullback explicitly we get $$ f^*\omega(\xi) = \alpha(e^{\mathrm{i}\pi/2}U\xi)\, \mathrm{det}(e^{\mathrm{i}\pi/2}U)\, \mathrm{d}\xi_1 \wedge \mathrm{d}\xi_2 = -\alpha(e^{\mathrm{i}\pi/2}U\xi)\, \mathrm{d}\xi_1 \wedge \mathrm{d}\xi_2 $$ noting the minus sign. So the result states that $$ \int_{-\infty}^\infty \int_{-\infty}^\infty \alpha(\xi)\, \mathrm{d}\xi_1 \, \mathrm{d}\xi_2 = -\int_{-\infty}^\infty \int_{-\infty}^\infty \alpha(e^{\mathrm{i}\pi/2}U\xi)\, \mathrm{d}\xi_1 \, \mathrm{d}\xi_2. $$ However, I've checked this on Mathematica for 1,000s of examples (for the class of functions $\alpha$ and $U\in \mathrm{SU}(2)$ I'm dealing with) and it seems that the minus sign shouldn't be there. Does anyone see the error in the previous analysis?