Isolate imaginary/real part of simple algebraic expression?

Posted 1 month ago
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 Using Mathematica can be often very frustrating. Either I'm doing something wrong here, or Mathematica is just dumb.So, I have the below expression: -1 + ((2*I*(1 + Log[Pi] - (I*Pi)/2))/Pi)*(r + I*t) And I am trying to put it in form a+i*b, nothing complicated at all, but Mathematica chokes on it. I tried to use the below, but it's ignoring my attempts to tell it that r and t are reals: In[298]:= Clear[r, t]; x1 = -1 + ((2*I*(1 + Log[Pi] - (I*Pi)/2))/ Pi)*(r + I*t); FullSimplify[Abs[x1]*Cos[Arg[x1]], Assumptions -> r*In*Reals && t*In*Reals] Out[298]= -1 - Im[t] + Re[r] - (2*(1 + Log[Pi])*(Im[r] + Re[t]))/Pi Where am I going wrong? Please help, I really appreciate.
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Posted 1 month ago
 Since r and t are both Real ComplexExpand[-1 + ((2*I*(1 + Log[Pi] - (I*Pi)/2))/Pi)*(r + I*t)] returns -1 + r - (2*t)/Pi - (2*t*Log[Pi])/Pi + I*((2*r)/Pi + t + (2*r*Log[Pi])/Pi) 
Posted 1 month ago
 Thanks, that's the function I needed.
Posted 1 month ago
 Wow, the result is absolutely horrible. Mathematica doesn't simplify a lot of the things it should: ComplexExpand[(Cos[r v] Cosh[t v] - I Sin[r v] Sinh[t v]) (Cos[t Log[Cos[v]]] - I Sin[t Log[Cos[v]]])] Cos[v] Cos[1/2 t Log[Cos[v]^2]] Cosh[t v] Cosh[t Arg[Cos[v]]] - Cosh[t Arg[Cos[v]]] Sin[v] Sin[1/2 t Log[Cos[v]^2]] Sinh[t v] + Cos[v] Cos[1/2 t Log[Cos[v]^2]] Cosh[t v] Sinh[t Arg[Cos[v]]] - Sin[v] Sin[1/2 t Log[Cos[v]^2]] Sinh[t v] Sinh[t Arg[Cos[v]]] + I (-Cos[v] Cosh[t v] Cosh[t Arg[Cos[v]]] Sin[1/2 t Log[Cos[v]^2]] - Cos[1/2 t Log[Cos[v]^2]] Cosh[t Arg[Cos[v]]] Sin[v] Sinh[t v] - Cos[v] Cosh[t v] Sin[1/2 t Log[Cos[v]^2]] Sinh[t Arg[Cos[v]]] - Cos[1/2 t Log[Cos[v]^2]] Sin[v] Sinh[t v] Sinh[t Arg[Cos[v]]]) 
 Maybe something like this? Assuming[Element[{v, t, r}, Reals] && Cos[v]^2 != 0, FullSimplify@PiecewiseExpand[ReIm@ComplexExpand[ (Cos[r v] Cosh[t v] - I Sin[r v] Sinh[t v])* (Cos[t Log[Cos[v]]] - I Sin[t Log[Cos[v]]])]]]