Complexification of $\mathfrak{so}(p,q)$ I have that

$\mathfrak{so}(p,q) = \{ A \in M_{p+q}(\mathbb{R}) : A^{T} I_{p,q} + I_{p,q} A = 0\}$. 
$\mathfrak{so}(p,q)_{\mathbb{C}} = \{ A \in M_{p+q}(\mathbb{C}) : A^{T} I_{p,q} + I_{p,q} A = 0\}$.
$\mathfrak{so}(p+q) = \{ A \in M_{p+q}(\mathbb{C}) : A^{T} + A = 0\}$.
I need to show that $\mathfrak{so}(p,q)_{\mathbb{C}} \cong \mathfrak{so}(p+q)$.

Here $I_{p,q} = \begin{bmatrix} I_{p} & 0 \\ 0 & -I_{q} \end{bmatrix}$. I was considering the map:

$\phi : \mathfrak{so}(p,q)_{\mathbb{C}} \rightarrow \mathfrak{so}(p+q)$, $A \mapsto I_{p,q} A$

It is bijective but not a lie algebra homomorphism, is there another mapping I should consider?
 A: Let's abbreviate $n:=p+q$. What you're using essentially is that over $\mathbb C$, the quadratic form defined by $I_{p,q}$ is congruent to the standard one defined by $I_{n}$. Find a base change matrix for that, i.e. $B \in GL_{n}(\mathbb C)$ such that $B^T \cdot I_{p,q} \cdot B = I_n$. (Hint: There will be imaginary entries, as the result we're getting at is not true over $\mathbb R$.) Then the same base change matrix will give an isomorphism of those Lie algebras, although now base change is of course given by $A \mapsto B^{-1} \cdot A \cdot B$. 
To show this $f(A):= B^{-1}AB$ indeed induces an isomorphism of the Lie algebras as defined in the OP, follow the hint in the comment. If you fail to see it, here:

 $$A \in \mathfrak{so}(p,q)_{\mathbb C}$$
$$\Leftrightarrow A^{T} I_{p,q} + I_{p,q} A = 0$$
 -- multiply both sides with $B^T$ from the left and $B$ from the right --
$$ \Leftrightarrow B^T A^{T} I_{p,q}B + B^T I_{p,q} AB = 0$$
 -- calling $B^{-T}$ the inverse of the transpose = transpose of the inverse, insert $I=B^{-T}B^T$ and $=BB^{-1}$ --
$$\Leftrightarrow \underbrace{B^T A^{T} B^{-T}}_{f(A)^T} \underbrace{B^T I_{p,q}B}_{I_n} + \underbrace{B^T I_{p,q} B}_{I_n}\underbrace{B^{-1}AB}_{f(A)} = 0$$
$$\Leftrightarrow f(A) \in \mathfrak{so}(p+q)$$

