Complex vector/matrix multiplication I am currently studying a textbook and I have a limited background in complex linear algebra and unfortunately the textbook does not provide background knowledge for this either. As a result I am a bit stuck with something that should be fairly simple. The problem is as follows:

Given known $3\times1$ vectors $\mathbf{h_1}$ and $\mathbf{h_2}$ we want to solve for an unknown matrix $C$ of shape $3\times3$ consisting of real entries. Suppose that we form a constraint with them such that $$(\mathbf{h_1} \pm i\mathbf{h_2})^TC(\mathbf{h_1} \pm i\mathbf{h_2})=0$$
Where $i\mathbf{h_2}$ denotes that vector $\mathbf{h_2}$ is an imaginary component. Then such constraint can be rewritten as two constraints:
$$\mathbf{h_1}^TC\mathbf{h_2}=0$$
$$\mathbf{h_1}^TC\mathbf{h_1}=\mathbf{h_2}^TC\mathbf{h_2}$$

Can anyone provide a thorough derivation of this? I suspect it is something to do with splitting up the real and imaginary parts or the plus minus sign?
 A: The second constraint can be easily shown by the separation of the real and imaginary parts (as you actually noted). I can prove the first constraint similarly but only for a symmetric C matrix (it is not clearly stated but I suppose it is symmetric from your comment).
I start by expanding the left-hand side:
$$ (\mathbf{h_1} \pm i\mathbf{h_2})^TC(\mathbf{h_1} \pm i\mathbf{h_2})=(\mathbf{h_1}^T \pm i\mathbf{h_2}^T)C(\mathbf{h_1} \pm i\mathbf{h_2}) = \mathbf{h_1}^TC\mathbf{h_1} \pm i\mathbf{h_1}^TC\mathbf{h_2} \pm i\mathbf{h_2}^TC\mathbf{h_1} + i^2\mathbf{h_2}^TC\mathbf{h_2} =  \mathbf{h_1}^TC\mathbf{h_1} \pm i\mathbf{h_1}^TC\mathbf{h_2} \pm i\mathbf{h_2}^TC\mathbf{h_1} - \mathbf{h_2}^TC\mathbf{h_2} = 0 $$
You need to realize that $ 0=0+0i $ and you can balance both real and imaginary part separately.
Balancing the real part gives you directly the second constraint:
$$ \mathbf{h_1}^TC\mathbf{h_1} - \mathbf{h_2}^TC\mathbf{h_2} = 0 \implies \mathbf{h_1}^TC\mathbf{h_1} = \mathbf{h_2}^TC\mathbf{h_2} $$
Balancing the imaginary part is a little bit more tricky:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_2}^TC\mathbf{h_1} = 0 $$
By using the property of transpose operation that $ (AB)^T=B^TA^T $ I get:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_2}^TC\mathbf{h_1} = \pm \mathbf{h_1}^TC\mathbf{h_2} \pm (\mathbf{h_1}^TC^T\mathbf{h_2})^T = 0 $$
Both terms are scalar, which can be proved simply by dimensionality analysis, and transpose of a scalar is the scalar itself:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm (\mathbf{h_1}^TC^T\mathbf{h_2})^T = \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_1}^TC^T\mathbf{h_2} = 0 $$
Now, you get finally your other constraint if $C$ is symmetric (i.e. $ C=C^T $):
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_1}^TC^T\mathbf{h_2} = \pm 2 \mathbf{h_1}^TC\mathbf{h_2} = 0 \implies \mathbf{h_1}^TC\mathbf{h_2} = 0 $$
