# CKKS Conjugation Let $m(X)$ be a plaintext, hence $m(\zeta_j)=z_j$ . The transformation: $m(X)\rightarrow m(X^{-1}) \bmod (X^n+1)$ is decoded to $\overline{z_j}$ because $\zeta_j^{-1}=\bar\zeta_j$ hence $m(\zeta_j^{-1})=m(\overline{\zeta_j}) = \overline{m(\zeta_j)} = \overline{z_j}$ . For $0< k \le n-1$ we have, $X^k\rightarrow {(X^k)}^{-1}=X^{-k}=-X^n\cdot X^{-k}=-X^{n-k}$ and for $k=0$, $X^0\rightarrow (X^0)^{-1}=X^0$. Reference: [[@Bootstrapping for Approximate Homomorphic Encryption]], section 4.2, page 10. ## Created 2024-09-10 14:04