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# Simplifying complex expression of Exp or trig functions

Posted 10 years ago
 I want do get the imaginary part of Exp[I*t]/(1-Exp[I*t])  I've tried to do this Im[ComplexExpand[Exp[I*t]/(1-Exp[I*t])]]  but the result still contains Im[Cos[t]/((1 - Cos[t])^2 + Sin[t]^2) - Cos[t]^2/((1 - Cos[t])^2 + Sin[t]^2) - Sin[t]^2/((1 - Cos[t])^2 + Sin[t]^2)]  which in fact is 0 (t is a real variable). How can I convince Mathematica to get rid of this part of the expression?
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Posted 10 years ago
 The last simplification even works without the assumptions: Simplify[ComplexExpand[Im[Exp[I*t]/(q - Exp[I*t])]]] gives (q Sin[t])/(1 + q^2 - 2 q Cos[t]) 
Posted 10 years ago
 Put the Im inside the ComplexExpand. In:= ComplexExpand[Im[Exp[I*t]/(1-Exp[I*t])]] Out= Sin[t]/((1-Cos[t])^2+Sin[t]^2) In:= Simplify[ComplexExpand[Im[Exp[I*t]/(q-Exp[I*t])]],Element[{q,t},Reals]] Out= (q Sin[t])/(1+q^2-2 q Cos[t]) Better?
Posted 10 years ago
 What I reallyn wnatm though, is Simplify[Im[ComplexExpand[Exp[I*t]/(q - Exp[I*t])]], Element[{q, t}, Reals]] which produces (-1 + q Cos[t]) Im[1/(1 + q^2 - 2 q Cos[t])] + q Re[1/(1 + q^2 - 2 q Cos[t])] Sin[t] but with real t and q Im[1/(1 + q^2 - 2 q Cos[t])] is 0 and  1/(1 + q^2 - 2 q Cos[t])] is real, therefore Re in the second part of the expression is unneccessaryBut what I really need id the following: Simplify[Im[ComplexExpand[Exp[I*t]/(q - Exp[I*t])]], {Element[{q, t}, Reals], Abs[q] < 1}] It gives the answer I expect: (q Sin[t])/(1 + q^2 - 2 q Cos[t]) 
Posted 10 years ago
 Re[...] should not be necessary here! When t is real, Cot[t/2] also is Real But Simplify[Im[ComplexExpand[Exp[I*t]/(1 - Exp[I*t])]], Element[t, Reals]] does the trick!
Posted 10 years ago
 Hi, I am not sure whether I fully understand the question, but does Simplify do the trick?  In= Im[ComplexExpand[Exp[I*t]/(1 - Exp[I*t])]] // Simplify Out= 1/2 Re[Cot[t/2]] Cheers, Marco