string theory - Expressions of action and energy momentum tensor in bc conformal field with central charge equals one

I have a question with conformal field theory in Polchinski's string theory vol 1 p. 51.

For $bc$ conformal field theory $$ S=\frac{1}{2\pi} \int d^2 z b \bar{\partial} c $$ $$ T(z)= :(\partial b) c: - \lambda \partial (: bc:) $$ with central charge $c=-3 (2 \lambda-1)^2 +1 =1$. Introducing $\psi$ and $\bar{\psi}$ to replace the anticommuting fields $b$ and $c$ as following $$b \rightarrow \psi =2^{-1/2} (\psi_1 + i \psi_2 )$$ and $$ c \rightarrow \bar{\psi} =2^{-1/2} (\psi_1 - i \psi_2)$$ It is claimed that $$S=\frac{1}{4\pi} \int d^2 z \psi_1 \bar{\partial} \psi_1 + \psi_2\bar{\partial} \psi_2 (2.5.18b)$$ $$ T=- \frac{1}{2} \psi_1 \partial \psi_1 -\frac{1}{2} \psi_2 \partial \psi_2 (2.5.18c)$$

I cannot obtain the above expressions of $S$ and $T$. Here is my derivations. First I try to recover the anti-commuting characters of fields $b$ and $c$ by $\psi_1$ and $\psi_2$. For $$bc+cb=0$$ I have $$\psi_1 \psi_1 + \psi_2 \psi_2 =0$$ Then for the action $$S=\frac{1}{4\pi} \int d^2 z \left( \psi_1 \bar{\partial} \psi_1 + i \psi_2 \bar{\partial} \psi_1 -i \psi_1 \bar{\partial} \psi_2 + \psi_2 \bar{\partial} \psi_2 \right)$$ (1) [Solved] Why the term $i \psi_2 \bar{\partial} \psi_1 -i \psi_1 \bar{\partial} \psi_2$ does not contribute to the action?

(2) How to derive (2.5.18c)?