Searching for elegant way manipulate complex vectors

Question

I have some complex vectors that I would like to normalize and further manipulate. For example, one is this:

${\qquad \rm vec}=(-\frac{-a + b + \sqrt{a^2 - 2 a b + b^2 + 4 c^2 + 4 d^2}}{2 (c + i d)},1)$

where I have declared $a,b,c,d$ to be real.

I normalized it simply by doing FullSimplify[Normalize[vec]], which resulted in the following normalization factor:

$\qquad \frac{1}{2}\sqrt{4 + {\rm Abs}(\frac{-a + b +\sqrt{(a - b)^2 + 4 (c^2 + d^2)}}{c + i d})^2}$

I do not want the absolute value in this expression. I want to to be:

$\qquad \frac{1}{2}\sqrt{4 + \frac{(-a + b +\sqrt{(a - b)^2 + 4 (c^2 + d^2)})^2}{c^2 + d^2}}$

My first question: is there an elegant way to produce this normalization factor?

There is an additional complication. If I hard-code in this normalization factor, call it normFactor, and attempt to take the conjugate of my normalized vector like so:

FullSimplify[Conjugate[vec/normFactor]]

The result has Conjugate in it. For example one of the vector components is

$\qquad 2{\rm Conjugate}\frac{1}{\sqrt{4+\frac{(-a+b+\sqrt{(a - b)^2 + 4 (c^2 + d^2)})^2}{c^2 + d^2}}}$

even though the term under the square roots are all positive because I have declared $a,b,c,d$ to be real.

My second and third questions: why is Conjugate behaving this way, and how can I fix it?

Answer 1

$Assumptions = a ∈ Reals && b ∈ Reals && c ∈ Reals && d ∈ Reals

vec = {-((-a + b + Sqrt[a^2 - 2 a b + b^2 + 4 c^2 + 4 d^2])/(2 (c + I d))), 1}

Norm[vec] // FullSimplify // ComplexExpand // FullSimplify

(*1/2 Sqrt[(Sqrt[(a - b)^2 + 4 (c^2 + d^2)] - a + b)^2/(c^2 + d^2) + 4]*)