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Sign error corrected, result reduced to something readable
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Roland F
Roland F
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Better forget Mathematicas Complex simplifying algorithms. For purely numerical evaluations the shortest way is direct manaipulationmanipulation by the operations definitions. At least one knows what one is doing:
(Just indent your code to get it pretty printed and copyable text, never use subscripts in communications)

 Subscript[z, 1]z1 = Sqrt[5 - Sqrt[3 I]]; Subscript[z,z2 2]= =PiPi/I - I/Pi; Subscript[z, 3]z3 = Exp[I] - I^E;

Now we calculate using a replacement rule for complec numbers

 
  Timing[(res0 = ComplexExpand[((Subscript[z, 1]z1 - Subscript[z, 2]z2 + Subscript[z, 3]z3) /. 
  Complex[a_, b_] :> 
   a - I b)/((Sqrt[#*(# /. Complex[a_, b_] :> a - I b)/] &)[
  z1 + z2 - Sqrt[(Subscript[z,z3])];



LeafCount[res0] 1] 
 +  Subscript[z, 2]8973 -
     N[res0]
    Subscript[z, 3])*0.566439 - 1.03623 I

Now we calculate using a replacement rule for complex numbers

res=((Subscript[z, 1]z1 - Subscript[z, 2]z2 + Subscript[z, 3]z3) /. 
 Complex[a_, b_] :> a - I b)]) // FullSimplify]
((Sqrt[#*(# /.

It takes a rather long time

Complex[a_, b_] :> {20.8906,a Sqrt[(-I +I \[Pi]b)] &)[z1+z2-z3])

(E^-(-I )^E+Sqrt[5- E^(-1)^(1/24) Sqrt[3]]+
E^-I E-I/\[Pi]-I \[Pi]) /
(\[Sqrt]((I^E+Sqrt[5-(-1)^(1/4) Sqrt[3]]-E^I-I/\[Pi]-
I \[Pi]) +((-I)^E+Sqrt[5-(-1)^(1/4) Sqrt[3]]-E^-
I+I/\[Pi]+I Sqrt[5\[Pi])))

N[res]

0.566439 +- Root[91.03623 +I

Finally one tries to get a normal algebraic expressions in many steps

ToRadicals@ #^4&FullSimplify[
 ,Together@ 1,ExpandAll[res] 0]]))/(-I. + 
 \[Pi]{Power[a_Complex (-E^I| +a_?Negative, E^((
b_] I:> EExp[b \[Pi])/2)Log[a]]}]

$$\frac{e^{-\frac{1}{2} i (2+e \pi )} \left(\frac{1}{2} e^{\frac{i e \pi }{2}} \left(2 \pi +e^i \left(\left(\sqrt{20-(2+2 i) \sqrt{6}}-2 i \pi \right) \pi -2 i\right)\right)-e^i \pi \right)}{\sqrt{1+\pi \left(2 \left(\sin (1)-\sin \left(\frac{e \pi }{2}\right)\right)+\pi \left(-\sqrt[4]{-1} \sqrt{3}+9+\pi \left(\pi +2 \sin (1)-2 \sin \left(\frac{e \pi }{2}\right)\right)-2 \cos \left(1-\frac{e \pi }{2}\right)+\sqrt{20-(2+2 i) \sqrt{6}} \left(\cos \left(\frac{e \pi }{2}\right)-\cos (1)\right)\right)\right)}}$$

 - ILeafCount[res2]
 \[Pi] +169
 Sqrt[5 + Root[9 + #^4& ,

N[res2]
0.566439 - 1,.03623 0]]))]}I

,

Better forget Mathematicas Complex simplifying algorithms. For purely numerical evaluations the shortest way is direct manaipulation by the operations definitions. At least one knows what one is doing:
(Just indent your code to get it pretty printed and copyable text)

 Subscript[z, 1] = Sqrt[5 - Sqrt[3 I]]; Subscript[z, 2] =Pi/I - I/Pi; Subscript[z, 3] = Exp[I] - I^E;

Now we calculate using a replacement rule for complec numbers

  Timing[(((Subscript[z, 1] - Subscript[z, 2] + Subscript[z, 3])/. 
    Complex[a_, b_] :> a - I b)/ 
      Sqrt[(Subscript[z, 1] +  Subscript[z, 2] - 
    Subscript[z, 3])*((Subscript[z, 1] - Subscript[z, 2] + Subscript[z, 3]) /. 
 Complex[a_, b_] :> a - I b)]) // FullSimplify]

It takes a rather long time

   {20.8906, Sqrt[(-I + \[Pi] (E^-I - E^(-(1/2) I E \[Pi]) - I \[Pi] + 
 Sqrt[5 + Root[9 + #^4& , 1, 0]]))/(-I + \[Pi] (-E^I + E^((
 I E \[Pi])/2) - I \[Pi] + Sqrt[5 + Root[9 + #^4& , 1, 0]]))]}

Better forget Mathematicas Complex simplifying algorithms. For purely numerical evaluations the shortest way is direct manipulation by the operations definitions. At least one knows what one is doing:
(Just indent your code to get it pretty printed and copyable text, never use subscripts in communications)

z1 = Sqrt[5 - Sqrt[3 I]]; z2 = Pi/I - I/Pi; z3 = Exp[I] - I^E;
 
  res0 = ComplexExpand[((z1 - z2 + z3) /. 
  Complex[a_, b_] :> 
   a - I b)/((Sqrt[#*(# /. Complex[a_, b_] :> a - I b)] &)[
  z1 + z2 - z3])];



LeafCount[res0]  
    8973 
     N[res0]
     0.566439 - 1.03623 I

Now we calculate using a replacement rule for complex numbers

res=((z1 - z2 + z3) /.Complex[a_, b_] :> a-I b)/
((Sqrt[#*(# /. Complex[a_, b_] :> a - I b)] &)[z1+z2-z3])

(-(-I)^E+Sqrt[5-(-1)^(1/4) Sqrt[3]]+
E^-I-I/\[Pi]-I \[Pi])/
(\[Sqrt]((I^E+Sqrt[5-(-1)^(1/4) Sqrt[3]]-E^I-I/\[Pi]-
I \[Pi]) ((-I)^E+Sqrt[5-(-1)^(1/4) Sqrt[3]]-E^-
I+I/\[Pi]+I \[Pi])))

N[res]

0.566439 - 1.03623 I

Finally one tries to get a normal algebraic expressions in many steps

ToRadicals@ FullSimplify[
 Together@ ExpandAll[res] /.  
 {Power[a_Complex | a_?Negative, b_] :> Exp[b Log[a]]}]

$$\frac{e^{-\frac{1}{2} i (2+e \pi )} \left(\frac{1}{2} e^{\frac{i e \pi }{2}} \left(2 \pi +e^i \left(\left(\sqrt{20-(2+2 i) \sqrt{6}}-2 i \pi \right) \pi -2 i\right)\right)-e^i \pi \right)}{\sqrt{1+\pi \left(2 \left(\sin (1)-\sin \left(\frac{e \pi }{2}\right)\right)+\pi \left(-\sqrt[4]{-1} \sqrt{3}+9+\pi \left(\pi +2 \sin (1)-2 \sin \left(\frac{e \pi }{2}\right)\right)-2 \cos \left(1-\frac{e \pi }{2}\right)+\sqrt{20-(2+2 i) \sqrt{6}} \left(\cos \left(\frac{e \pi }{2}\right)-\cos (1)\right)\right)\right)}}$$

  LeafCount[res2]
  169
      

N[res2]
0.566439 - 1.03623 I

,

Source Link
Roland F
Roland F
  • 4.6k
  • 1
  • 2
  • 12

Better forget Mathematicas Complex simplifying algorithms. For purely numerical evaluations the shortest way is direct manaipulation by the operations definitions. At least one knows what one is doing:
(Just indent your code to get it pretty printed and copyable text)

 Subscript[z, 1] = Sqrt[5 - Sqrt[3 I]]; Subscript[z, 2] =Pi/I - I/Pi; Subscript[z, 3] = Exp[I] - I^E;

Now we calculate using a replacement rule for complec numbers

  Timing[(((Subscript[z, 1] - Subscript[z, 2] + Subscript[z, 3])/. 
    Complex[a_, b_] :> a - I b)/ 
      Sqrt[(Subscript[z, 1] +  Subscript[z, 2] - 
    Subscript[z, 3])*((Subscript[z, 1] - Subscript[z, 2] + Subscript[z, 3]) /. 
 Complex[a_, b_] :> a - I b)]) // FullSimplify]

It takes a rather long time

   {20.8906, Sqrt[(-I + \[Pi] (E^-I - E^(-(1/2) I E \[Pi]) - I \[Pi] + 
 Sqrt[5 + Root[9 + #^4& , 1, 0]]))/(-I + \[Pi] (-E^I + E^((
 I E \[Pi])/2) - I \[Pi] + Sqrt[5 + Root[9 + #^4& , 1, 0]]))]}