= Solution
The <complexification of a Lie algebra> $L_0$ is $L=L_0\otimes_{\mathbb R}\mathbb C$, with the <Lie bracket> extended complex-bilinearly. Equivalently write $L=L_0\oplus iL_0$, where
$$
[x+iy,z+iw]=[x,z]-[y,w]+i([x,w]+[y,z]).
$$
Complex conjugation is an <antilinear map> and a <Lie algebra automorphism> whose fixed subalgebra is $L_0$.
If the <solvable radical> of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the <solvable radical> $R$ of $L$ is preserved by <complex conjugation>, because it is the unique largest solvable ideal. Consequently
$$
R=R_0\oplus iR_0,\qquad R_0=R\cap L_0:
$$
for $z\in R$, both $(z+\overline z)/2$ and $(z-\overline z)/(2i)$ belong to $R_0$. If $R\ne0$, $R_0\ne0$ is a solvable ideal of $L_0$. This proves \b[$L_0$ is semisimple if and only if $L_0\otimes_{\mathbb R}\mathbb C$ is semisimple]. Consistently, the complex <Killing form> is just the complex-bilinear extension of the real one; its determinant in a real <basis> is unchanged by extending scalars.
Take the <sl2R Lie algebra> $\mathfrak{sl}_2(\mathbb R)$ and the <special unitary Lie algebra> $\mathfrak{su}(2)$. Both complexify to $\mathfrak{sl}_2(\mathbb C)$. This is immediate for the former; for the latter, the real <basis> $ih,e-f,i(e+f)$ consists of <traceless matrices> that are <skew-Hermitian matrices> and is also a complex <basis> of $\mathfrak{sl}_2(\mathbb C)$.
They are not isomorphic as real <Lie algebras>. Their <Killing forms> are $B(X,Y)=4\operatorname{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On $\mathfrak{sl}_2(\mathbb R)$ the <basis> $h,e+f,e-f$ has a diagonal <Gram matrix> with entries $8,8,-8$, so the <signature of a quadratic form> is $(2,1)$. A <Lie algebra isomorphism> preserves the <Killing form> and therefore its <signature>.
A real <split semisimple Lie algebra> has a <Cartan subalgebra> whose <Adjoint representation of a Lie algebra> is simultaneously diagonalizable over $\mathbb R$, so its <root-space decomposition> is defined over $\mathbb R$. The example $\mathfrak{sl}_2(\mathbb R)$ is split: the diagonal <Cartan subalgebra> $\mathbb Rh$ has <eigenvalues> $0,2,-2$ and real <root spaces> $\mathbb Re,\mathbb Rf$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\operatorname{ad}x$ skew-adjoint for the positive definite <inner product> $-B$, so its <eigenvalues> are purely imaginary. If it were diagonalizable over $\mathbb R$, all these eigenvalues would be zero and $\operatorname{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split <Cartan subalgebra> exists. Thus \b[$\mathfrak{sl}_2(\mathbb R)$ is split and $\mathfrak{su}(2)$ is not].
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