Inner product of imaginary vectors? 
I'm blanking on how to do the inner product of imaginary vectors. Would the imaginary part just become negative? Also how do we show those inequalities hold. This second part is the main problem I have.
 A: Read the book before this point.  My guess: the inner product is defined with a conjugate in there, so
$$
\langle u,  v\rangle =(1)\overline{(1+i)}+(i)\overline{(2+3i)}+(1+i)\overline{(4+5i)}+(0)\overline{(6+7i)}
$$
This way, we always get $\langle x, x \rangle \ge 0$, so taking its square-root is a sensible way to compute the norm.
But if it is a physics book, the conjugate will be on the first factor instead of the second factor.
A: For the inner product, would you not just do it as you normally would? Ie, 
\begin{equation*}
u\cdot v=(1)(1+i)+(i)(2+3i)+(1+i)(4+5i)+(0)(6+7i)\\
\end{equation*}
and expand watching out for the complex multiplication and double brackets (ie, $xy=(ac-bd)+i(ad+bc)$ for complex numbers $x=a+ib$ and $y=c+id,~i^2=-1$)?
For the second part, why not try substituting the vectors into the equations and see what you get? Ie, into
\begin{equation*}
||u+v||\leq ||u||+||v||,\\
|\langle u,v \rangle|\leq ||u||~||v||.
\end{equation*}
A: $\langle v , w \rangle= \langle (1,i,1+i,0),(1+i,2+3i,4+5i,6+7i)\rangle$
the only thing we really need to recall here is that $i=\sqrt{-1}$ that is $i^2=-1$
$\langle v , w \rangle =(1)(1+i)+(i)(2+3i)+(1+i)(4+5i)+(0)(6+7i)$
$=1+i+2i+3i^2+4+5i^2$
$=5+3i-3-5$
$=3i-3$
$||v||=\sqrt{1+i^2+(1+i)^2}$
and $||w||= \sqrt{(1+i)^2+(2+3i)^2+(4+5i)^2+(6+7i)^2}$
to verify the inequalities, just use the definitions
