of the corresponding operators evaluated between a $B$ meson and the $\pi K$ isospin state with $I=\frac32$. Since our intention here is to look for New Physics effects rather than doing precision calculations, it will be sufficient for our purposes to evaluate these matrix elements in a given New Physics scenario using the naive factorization approximation and neglecting small SU(3)-breaking effects.[^3] Then the strong-interaction phases $\phi_a$ and $\phi_b$ vanish, and we obtain $$\begin{aligned}
\label{abres}
   |\lambda_u|\,(a+ib) &\approx&
    - \left( \bar c_3^{\rm EW} + \bar c_4^{\rm EW} \right) 
    + \frac23\,\kappa^{3/23}
    \left( \bar c_1^{\rm EW} + \bar c_5^{\rm EW} \right)
    \nonumber\\
   &&\mbox{}- \frac{3\chi-1}{4}\,\kappa^{30/23}
    \bigg( \bar c_2^{\rm EW} + \frac13\,\bar c_1^{\rm EW} 
    + \bar c_6^{\rm EW} + \frac13\,\bar c_5^{\rm EW} \bigg) \,,\end{aligned}$$ where $\bar c_i^{\rm EW}\equiv c_i^{\rm EW}(m_W)
-\widetilde c_i^{\rm EW}(m_W)$, and $$\chi = \frac{2 m_K^2}{(m_s+m_d)\,m_b}
   = \frac{2 m_\pi^2}{(m_u+m_d)\,m_b} \,.$$ Parity invariance implies that the relevant hadronic matrix elements of the operators $\widetilde O_i$ have the opposite sign compared with those of the operators $O_i$. The fact that $\bar c_3^{\rm EW}$ and $\bar c_4^{\rm EW}$ enter with the coefficient $-1$ in (\[abres\]) is a model-independent result free of hadronic uncertainties, irrespective of whether these coefficients receive New Physics contributions or not. It follows because 