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simplifying expressions - Separating exponential terms


Suppose I have a lot of expressions multiplied by factors such as:


e−iθ[1]−iθ[2]−iθ[3]−iθ[4]−iθ[5]


I would like to separate this into a product of exponentials of the form


e−iθ[1]e−iθ[2]...


before employing the function ExpToTrig and making substitutions to the result trigonometric functions.



However, since I plan to apply the tangent half angle substitution (cf. my previous question Simplifying Expressions for FindMinimum), I would like the arguments to involve only one variable at a time. In particular, I tried using ComplexExpand on the function to express the trigonometric functions as functions of a single variable, but it expands the entire function out.


In short, I would like to keep the simplified form, but want to expand the exponential as per the above without having to expand the entire expression.


For reference, here is my function


(E^(-I θ[1] - I θ[2] - I θ[3] - I θ[4] - I (θ[5] - θ[6]))
Abs[Sin[ϕ[6]]]^2 (1 - E^(I (θ[1] - θ[6]))
   Cot[ϕ[6]/2] Tan[ϕ[1]/2]) (Cos[θ[1]] + I Sin[θ[1]] + E^(I θ[2])
   Tan[ϕ[1]/2] Tan[ϕ[2]/2]) (Cos[θ[2]] + I Sin[θ[2]] + E^(I θ[3])
   Tan[ϕ[2]/2] Tan[ϕ[3]/2]) (Cos[θ[3]] + I Sin[θ[3]] + E^(I θ[4])
   Tan[ϕ[3]/2] Tan[ϕ[4]/2]) (Cos[θ[5] - θ[6]] + I Sin[θ[5] - θ[6]] - Cot[ϕ[6]/2] 
   Tan[ϕ[5]/2]) (Cos[θ[4]] + I Sin[θ[4]] + E^(I θ[5])

   Tan[Ï•[4]/2] Tan[Ï•[5]/2]))/
(2 Sqrt[(1 + Abs[Tan[Ï•[1]/2]]^2) (1 + Abs[Tan[Ï•[2]/2]]^2)]
   Sqrt[(1 + Abs[Tan[Ï•[2]/2]]^2) (1 + Abs[Tan[Ï•[3]/2]]^2)]
   Sqrt[(1 + Abs[Tan[Ï•[3]/2]]^2) (1 + Abs[Tan[Ï•[4]/2]]^2)]
   Sqrt[(1 + Abs[Tan[Ï•[4]/2]]^2) (1 + Abs[Tan[Ï•[5]/2]]^2)]
   Sqrt[(1 + Abs[Tan[Ï•[1]/2]]^2) (1 + Cos[Ï•[6]])]
   Sqrt[(1 + Abs[Tan[Ï•[5]/2]]^2) (1 + Cos[Ï•[6]])])

Answer



Update


In the interest of simplifying the code somewhat, I've modified one of the replacements. For instance, we can do



expr2 = Thread[expr1, Plus] /. Plus -> Times

or


epxr2 = expr1 /. expT[Plus[a__]] :> Times @@ expT /@ a

rather than


expr2 = expr1 //. {expT[a_ + b_] :> expT[a] expT[b]}

So:


f[expr_] := Thread[expr /. Power[E, a_] :> expT@Expand@a, Plus] /. Plus -> Times /. expT[a_] :> ExpToTrig@Exp@a


or


f[expr_] := expr /. Power[E, a_] :> expT@Expand@a /. expT[Plus[a__]] :> Times @@ expT /@ a /. expT[a_] :> ExpToTrig@Exp@a

Original Post


As Bill commented, Mathematica likes to keep Exp[]'s together. Here's a workaround that I've used in the past. We replace Exp with a dummy head expT, do the re-write using replacement rules, and in the process apply ExpToTrig.


For instance, if


expr = Exp[-I (4 + a) + c];

we first do



expr1 = expr /. Power[E, a_] :> expT@Expand@a
(* expT[-4 I - I a + c] *)

Then, we separate the terms inside expT using ReplaceRepeated:


expr2 = expr1 //. {expT[a_ + b_] :> expT[a] expT[b]}
(* expT[-4 I] expT[-I a] expT[c] *)

Finally, we convert back to Exp and apply ExpToTrig:


expr2 /. expT[a_] :> ExpToTrig@Exp@a
(* (Cos[4] - I Sin[4]) (Cos[a] - I Sin[a]) (Cosh[c] + Sinh[c]) *)


We can do all at once, of course. Define


f[expr_] := expr /. Power[E, a_] :> expT@a //. {expT[a_ + b_] :> expT[a] expT[b]} /. expT[a_] :> ExpToTrig@Exp@a

in which case


f[expr]
(* (Cos[4] - I Sin[4]) (Cos[a] - I Sin[a]) (Cosh[c] + Sinh[c]) *)

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