Mathematica Asked on March 25, 2021
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?
$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]*)
Answered by Bill Watts on March 25, 2021
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