Given a Hermitian positive semidefinite matrix $A \in \mathbb{C}^{n \times n}$.
If $B=A^{-1},\ D=\text{Re}(B),\ C=D^{-1}$.
where $D=\text{Re}(B)\Leftrightarrow{}d_{ij}=\text{Re}(b_{ij})$.
Can I calculate matrix $C$ from $A$ directly without calculating matrix inverse twice (or any methods that require lower operations than this)?
We have $\operatorname{Re}(B) = \frac 12 (B + \bar B)$. Note that $\bar B = \overline{A^{-1}} = \bar A^{-1}$, so that $$ A(B + \bar B) \bar A = AB\bar A + A\bar B\bar A = \bar A + A = 2\operatorname{Re}(A). $$ That is, we have $$ 2A \operatorname{Re}(B) \bar A = 2\operatorname{Re}(A) \implies \\ \operatorname{Re}(B) = A^{-1}\operatorname{Re}(A)\bar A^{-1} \implies \\ \operatorname{Re}(B)^{-1} = \bar A \operatorname{Re}(A)^{-1} A. $$