Assuming $T=R_1+iS_1$, with $R_1$ and $S_1$ self-adjoint, for $u,v\in V$ we have $ \langle S_1(u),v\rangle=\langle -iT(u)+iR_1(u),v\rangle =-i\langle T(u),v\rangle +i\langle R_1(u),v\rangle. $
Also $ \langle S_1(u),v\rangle=\langle u,S_1(v)\rangle=\langle u,-iT(v)+iR_1(v)\rangle=i\langle u,T(v)\rangle-i\langle u,R_1(v)\rangle=i\langle T^*(u),v\rangle-i\langle R_1(u),v\rangle. $ Thus $ 0=-i\langle T(u),v\rangle +i\langle R_1(u),v\rangle-i\langle T^*(u),v\rangle+i\langle R_1(u),v\rangle=\langle -iT(u)-iT^*(u)+2iR_1(u),v\rangle. $ In particular, $ 0=\langle i(2R_1-(T+T^*))(u),i(2R_1-(T+T^*))(u)\rangle, $ so $i(2R_1-(T+T^*))(u)=0.$ Therefore $ R_1=\frac{1}{2}(T+T^*). $
Proving $S_1$ is similar.