What are the rules for complex-component vectors and why?

Step 1
Given complex vectors
$v=(v_1,\cdots ,v_n)$
and
$w=(w_1,\cdots ,w_n)$,
their scalar (=dot) product is given by
$v\times w=\sum _{j=1}^n{v}_j{\overline{w}}_j=v_1{\overline{w}}_1+\cdots +v_n{\overline{w}}_n$
Why? Well, you want to generalize the usual dot product on $\mathbb{R}$, but you also want that
$v\times v\ge 0$
for all v, and the vector $(i,0,\cdots ,0)$ shows that you can't do without the conjugation. You might ask why, I don't care if
$v\times v\ge 0$
for all v. Most people do: the expression
$||v||=\sqrt{v\times v}$
should define a norm and $||v-w||$ a metric on ${\mathbb{C}}^n$, and taking square-roots of non-positve numbers (or even complex numbers) simply isn't well-defined. Note that
$||v||=\sqrt{v\times v}={\sqrt{\sum _{j=1}}}^n|v_j{|}^2$
Having settled this, let v be a non-zero vector. Its orthogonal complement
$U=v^{\mathrm{\perp }}=\{u\in {\mathbb{C}}^n:u\times v=0\}$
is the set of all vectors orthogonal to v. Since U is determined by the single linear equation
$u_1{\overline{v}}_1+\cdots +u_n{\overline{v}}_n=0$,
it is an $(n-1)$ -dimensional subspace of ${\mathbb{C}}^n$ Finding solutions is easily achieved using Gauss elimination, this will give you vectors
$u_1,\cdots ,u_{n-1}$
which you can make into an orthonormal basis of U using Gram-Schmidt (note that the notation
$<<u,v>>=u\times v$
is to be understood. The fact that you're working with PK\mathbb{C}\) and not with $\mathbb{R}$ is immaterial, just be careful to note that
$v\times (\lambda w)=\overline{\lambda }(v\times w)$
i.e., the dot product is conjugate-linear in the second variable.
Finally, in order to solve the equation
$w\times v=d$
simply take any
$u\in U=v^{\mathrm{\perp }}$
and put
$w=u+\frac{d}{v\times v}v$,
and note that
$w\times v=(u+\frac{d}{v\times v}v)\times v=(u\times v)+\frac{d}{v\times v}v\times v=0+d=d$

To take the length of a complex vector you need the squared magnitudes of the components. So the vector $(1+i,1-i)$ has length $\sqrt{|1+i{|}^2+|1-i{|}^2}=\sqrt{2+2}=2$ It is still true that the dot product of orthogonal vectors is zero in a complex space. So once you find a B parallel to A with $A\times B=d$ you can add any vector orthogonal to A to it and the dot product will not change. But when you take the dot product, remember to take the complex conjugate of the components of B.