= Solution
The <derived series> gives a direct test, with no structural theorem needed. By bilinearity and antisymmetry of the <Lie bracket>, every bracket in $\mathfrak h$ is a multiple of $z$, and $z=[x,y]$ actually occurs. Therefore
$$
\mathfrak h^{(1)}=[\mathfrak h,\mathfrak h]=\mathbb Cz,\qquad
\mathfrak h^{(2)}=[\mathbb Cz,\mathbb Cz]=0.
$$
Consequently this <Heisenberg Lie algebra> is a nonzero <solvable Lie algebra>. It is an <ideal of a Lie algebra> in itself, since $[\mathfrak h,\mathfrak h]\subseteq\mathfrak h$. Its <solvable radical> is therefore all of $\mathfrak h$, contradicting the first criterion in part (a). Hence
$$
\boxed{\mathfrak h\text{ is not semisimple}.}
$$
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