math problem loss REVIEWED

Math Problem with R,L,C,G PhL 6.22.14 Early version of what was developed in Appendix Q. But that Appendix Q assumed constant parameters, which I had to replace by a new Appendix Q allowing for ω variation of the parameters R,L,G,C. But that only affected large and small ω limits. The result just below still appear as is in current Appendix Q, getting the real and imaginary part of k(ω) in essence. (10.8.14) What is shown here: s ≡ = Re(s) = (R2+ω2L2)1/4 (G2+ω2C2)1/4 Im(s) = (R2+ω2L2)1/4 (G2+ω2C2)1/4 In simpler form, first define a ≡ (R2+ω2L2)1/4 (G2+ω2C2)1/4 dim(a) = R/m a2 → ω2LC as ω → ∞ c ≡ RG - ω2LC dim(c) = 1/m2 b ≡ = c/a2 dim(b) = 1 b→ - 1 as ω → ∞; 1-b → 2 so that Re(s) = a Im(s) = a s = a [+ j ] Verification of this solution s = a [+ j ] = a [+ j ] = [+ j ] If I draw a triangle, the radius is a which is in fact correct. But how do I show the angle? From the above I can see that sinφ = / = [ ]1/2 = cosφ = / = [ ]1/2 Now ansatz that c/a2 = cosθ, then consistent with φ = θ/2, but is the ansatz right? ************************************************* Consider q ≡ (R+jωL)(G+jωC) = (RG-ω2LC) + jω(LG+RC) This q has positive imaginary part and unknown sign real part, so this argument is a vector in the upper half plane. We can then write q = |q| ejθ |q|2 = | (R+jωL)(G+jωC) |2 = | (R+jωL)|2|(G+jωC) |2 = (R2+ω2L2) (G2+ω2C2) tan θ = ω(LG+RC) / (RG-ω2LC) |q| = θ = tan-1[ω(LG+RC)] / (RG-ω2LC)] Note that the tangent jumps from +∞ to -∞ but the angle θ is continuous through π/2. Note that: cosθ = (RG-ω2LC) / |q| = (RG-ω2LC)/ = c/a2 sinθ = ω(LG+RC) / |q| = ω(LG+RC)/ So now define s ≡ = We can then write s = |s| eiφ |s| = φ = θ/2 so then |s| = (R2+ω2L2)1/4 (G2+ω2C2)1/4 = a φ = (1/2) tan-1[ω(LG+RC)] / (RG-ω2LC)] Now try to write out the real and imaginary parts of this: Re(s) = |s| cosφ = |s| cos(θ/2) Im(s) = |s| sinφ = |s| sin(θ/2) Now use cos φ = cos(θ/2) = = = = = Note that the magnitude of the big fraction is always less than 1, so everything is real. Similarly sinθ/2 = = / = Then here is my conclusion: s ≡ = Re(s) = (R2+ω2L2)1/4 (G2+ω2C2)1/4 Im(s) = (R2+ω2L2)1/4 (G2+ω2C2)1/4 If I ask Maple to do this, I get but then Maple does not realize that, if you take the positive root of the inner root, then the argument of each outer root is positive and you get my answer. Maple is worried you might take the negative root for the inner root. That is why we get and it refuses to take the real part.