The exact operator-norm epsilon region for determinant-one Rz candidates

Put $\mathrm{\Delta }=R_z(\theta {)}^{\ast }U$. It is unitary and has determinant $1$, so [F2] diagonalizes it. Its eigenvalues $\lambda ,\mu$ have modulus one by unitarity and satisfy $\lambda \mu =1$ by the determinant, hence $\mu =\bar{\lambda }$. By [F4], write $\lambda =e^{it}$ with $t\in [0,2\pi )$; if $t>\pi$, swap $\lambda ,\mu$ and put $\alpha =2\pi -t$, while if $t\le \pi$ put $\alpha =t$. Thus the eigenvalues are $e^{i\alpha },e^{-i\alpha }$ for some $\alpha \in [0,\pi ]$. From the displayed matrices, $\mathrm{tr}\mathrm{\Delta }=e^{i\theta \mathrm{/}2}u+e^{-i\theta \mathrm{/}2}\bar{u}=2\mathrm{Re}(e^{i\theta \mathrm{/}2}u)$. Hence $\cos \alpha =\mathrm{Re}(e^{i\theta \mathrm{/}2}u)$.

Left multiplication by $R_z(\theta {)}^{\ast }$ preserves vector norms, so [F3] gives $\mathrm{\parallel }{R}_z(\theta )-U{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }I-\mathrm{\Delta }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Let $S$ be a unitary diagonalizer of $\mathrm{\Delta }$ from [F2]; conjugation by $S$ preserves the operator norm by [F3]. Using [F4] then gives $\mathrm{\parallel }I-\mathrm{\Delta }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2=\max (\mathrm{\mid }1-e^{i\alpha }{\mathrm{\mid }}^2,\mathrm{\mid }1-e^{-i\alpha }{\mathrm{\mid }}^2)=2-2\cos \alpha =2-2\mathrm{Re}(e^{i\theta \mathrm{/}2}u)$. Both the operator norm and $\epsilon$ are nonnegative, so squaring the bound is equivalent to the stated real-part inequality.