# Why isn't Simplify completely simplifying my expression when it has all the necessary info?

I have a complex variable, and I'm trying to multiply it by its Conjugate[], which should give me a real number. So, I'm currently doing:

r = Simplify[(b/a)*Conjugate[b/a], {L, k1, kreal, kimag} \[Element] Reals]


Which is giving me:

(E^(-2 I kreal L) (-1 + E^(2 (kimag + I kreal) L)) (E^(2 kimag L) -
E^(2 I kreal L)) (k1^2 + (kimag - I kreal)^2) (k1^2 + (kimag +
I kreal)^2))/((-E^(2 I kreal L) (-k1 + I kimag + kreal)^2 +
E^(2 kimag L) (k1 + I kimag + kreal)^2) (E^(2 kimag L) k1^2 -
Conjugate[
E^(2 kimag L) (kimag - I kreal) (-2 I k1 + kimag - I kreal) +
E^(2 I kreal L) (-k1 + I kimag + kreal)^2]))


(I apologize for the poor formatting, I don't know how to do it so it looks nice here without editing a ton of latex.)

Anyway, clearly this is not as simplified as it could be -- it knows that every variable in that expression is real and I multiplied a complex number by its complex conjugate, so the entire expression should be real and free of $i$'s or Conjugate[]'s, but it isn't.

What am I missing?

Thanks!

edit: I realized I should have included a and b. a is intentionally undefined, but that shouldn't matter because it divides out from b:

(a (-1 + E^(2 I k2 L)) (k1^2 - k2^2))/(-k1^2 + E^(2 I k2 L) k1^2 - 2 k1 k2 - 2 E^(2 I k2 L) k1 k2 - k2^2 + E^(2 I k2 L) k2^2)


Where k2 is then defined as:

k2 = Simplify[kreal + I*kimag, {kimag, kreal} \[Element] Reals];


I've also tried just doing Simplify[] to the Conjugate[] of (b/a), which is confusing because it still has a Conjugate[] in the output:

Simplify[Conjugate[b/a], {L, k1, kreal, kimag} \[Element] Reals


Gives:

(E^(-2 I kreal L) (-1 + E^(2 (kimag + I kreal) L)) (k1^2 + (kimag + I kreal)^2))/(E^(2 kimag L) (k1^2 - kimag^2 + kreal^2) - Conjugate[E^(2 I kreal L) (-k1 + I kimag + kreal)^2 - 2 I E^(2 kimag L) (k1 (kimag - I kreal) + kimag kreal)])


Why? It knows that all the variables inside that Conjugate[] are real, so it should just flip the signs of the $i$ terms.

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How are {L, k1, kreal, kimag}  related to {a, b}? – Dr. belisarius Jan 7 '14 at 20:24
@belisarius, Thank you for the response. I've edited my post above to define b. I suspect it's because at some point I'm not telling it that the right variables are Reals, but I'm not sure why it doesn't propagate forward. – YungHummmma Jan 7 '14 at 20:58
ad your apology: You can use TeXForm to automatically convert your equations. – shrx Jan 7 '14 at 22:18

I'm not sure I follow the full problem, but one approach is

ComplexExpand[(b/a)*Conjugate[b/a]]
b^2/a^2


which automatically assumes the variables are real-valued. When applied to the output of your first expression, you get something real:

b = (a (-1 + E^(2 I k2 L)) (k1^2 - k2^2))/(-k1^2 +
E^(2 I k2 L) k1^2 - 2 k1 k2 - 2 E^(2 I k2 L) k1 k2 - k2^2 +
E^(2 I k2 L) k2^2);
FullSimplify[ComplexExpand[(b/a)*Conjugate[b/a]]]


gives a real-valued output

(2 (k1^2 - k2^2)^2 Sin[k2 L]^2)/(k1^4 + 6 k1^2 k2^2 + k2^4 - (k1^2 - k2^2)^2 Cos[2 k2 L])


Alternatively, if I work with your first expression but use FullSimplify and ComplexExpand (instead of just Simplify)

FullSimplify[
ComplexExpand[(E^(-2 I kreal L) (-1 +
E^(2 (kimag + I kreal) L)) (E^(2 kimag L) -
E^(2 I kreal L)) (k1^2 + (kimag - I kreal)^2) (k1^2 + (kimag +
I kreal)^2))/((-E^(2 I kreal L) (-k1 + I kimag + kreal)^2 +
E^(2 kimag L) (k1 + I kimag + kreal)^2) (E^(2 kimag L) k1^2 -
Conjugate[
E^(2 kimag L) (kimag - I kreal) (-2 I k1 + kimag - I kreal) +
E^(2 I kreal L) (-k1 + I kimag + kreal)^2]))]]


then you also get a real-valued output (assuming that both kreal and kimag are real-valued).

((k1^4 + 2 k1^2 (kimag - kreal) (kimag + kreal) + (kimag^2 +
kreal^2)^2) (1 + E^(4 kimag L) -
2 E^(2 kimag L) Cos[2 kreal L]))/((kimag^2 + (k1 - kreal)^2)^2 +
E^(4 kimag L) (kimag^2 + (k1 + kreal)^2)^2 +
2 E^(2 kimag L) (-(k1^4 + (kimag^2 + kreal^2)^2 -
2 k1^2 (3 kimag^2 + kreal^2)) Cos[2 kreal L] +
4 k1 kimag (-k1^2 + kimag^2 + kreal^2) Sin[2 kreal L]))

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Hi, thank you. I tried that before and got something insanely long. But also, in the output you have there, there is still I's and Conjugate[]'s in it -- I know they can cancel out to give a real number, but why are they even there if it can easily be simplified to not have them? – YungHummmma Jan 7 '14 at 21:24
It wasn't completely clear if this is the proper value for b, but if it is, there are no I's or conjugates in the output. – bill s Jan 7 '14 at 21:27
Yep, I think that's what I get now! – YungHummmma Jan 7 '14 at 21:46