Wedge product equality of 2n-forms in 2n+2 dimension

Question

I have a question that seems to me clear but i couldn't prove it. I have a diffeomorphism between the cotangent bundle of the n-sphere and the complex quadric as $$T^*(S^n)\to (S^n)^{\mathbb{C}},\ (x,p)\mapsto \cosh(|p|)x+i\frac{\sinh(|p|)}{|p|}p$$ where $(x,p)\in \mathbb{R}^{n+1}\times \mathbb{R}^{n+1}$ such that $|x|=1$ and $\langle p,x\rangle=0$, and $|p|$ is the usual euclidean norm. I have the canonical symplectic form $\omega=\sum_j^{n+1} dp_j\wedge dx_j$ on $T^*(S^n)$ who gives us the Liouville volume form $\varepsilon=\omega^{\wedge n}$ (I omit some constant coefficients). I have a volume form $\Omega=\sum_j^{n+1}(-1)^j z_j dz_1\wedge\ldots\wedge \widehat{dz_j}\wedge\ldots\wedge dz_{n+1}$ on $(S^n)^{\mathbb C}$ where $\widehat{dz_j}$ means that we forget this term. I know that $\omega^{\wedge n}$ is a 2n-form on $\mathbb R^{2n+2}$ and $\Omega\wedge\overline{\Omega}$ is again a 2n-form on $\mathbb R^{2n+2}$. My goal is to find the next function $$ \Omega\wedge\overline{\Omega}|_{T^*{S^n}}=f\cdot \omega^{\wedge n}|_{T^*{S^n}}.$$ But the calculations of 2n-forms in 2n+2 dimension is very difficult. I used the constraint of the complex quadric $(S^n)^{\mathbb C}$, i.e. $\sum_j^{n+1}z_j^2$ (who is equal to 1 on $(S^n)^{\mathbb C}$) and i did the calculations $$ \Omega\wedge\overline{\Omega}\wedge d(\sum_j^{n+1}z_j^2)\wedge \overline{d(\sum_j^{n+1}z_j^2)} |_{T^*{S^n}}=f_1\cdot \omega^{\wedge n}\wedge d(\sum_j^{n+1}z_j^2)\wedge \overline{d(\sum_j^{n+1}z_j^2)} |_{T^*{S^n}}.$$ I found $f_1$ easily in this way, but i have to show that $f=f_1$. Can anyone help me about this please?