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Let $(B,ω')$ be a symplectic Lie algebra. and the $\delta$ is a symplectic derivation, and $z\in B$.

The double extension of $B$ is: $g=ℝe ⊕ B ⊕ ℝd$ as:

Central Extension : $I = ℝe ⊕ B$,of $B$ with ℝe, and the bracket is : $[a, b]_I = [a, b]_B + ω'(δa, b)e$ pour $a, b ∈ B.$

Semi-direct Product : $I$ with $ℝd$, with the brackets : $[d, e] = 0,$ $[d, a] = −ω'(z, a)e − δ(a)$ for $ a ∈ B.$

My question is how we find the z?

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This is explained here, pages $17-22$. Since $\omega$ is nondegenerate there exists a unique $z=z_D\in B$ such that $f(x)=\omega(z,x)$ for all $x\in B$. Here $f([x,y])=\Omega_D(x,y)$ is a $2$-coboundary, see page $20$.

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