How to transform a complex quotient involving absolute values and conjugates into the rectangular form?

Question

I am relatively new to Mathematica and I am still trying to figure out all the commands. I have these three complex numbers:

And I wish to calculate the following quotient and write it in the a+bi form:

What I have done on Mathematica is the following:

I have been trying to figure out how to transform the last answer to the desired form but I can't figure it out. Maybe I have constructed the questions in a faulty manner but I simply don't know a better way. I would appreciate all the help I could get.

Answer 1

I think ComplexExpand gives the answer

z1 = Sqrt[5 - Sqrt[3 I]];
z2 = Pi/I - I/Pi;
z3 = Exp[I] - I^E;
z = Conjugate[z1 - z2 + z3]/Abs[z1 + z2 - z3];

ComplexExpand[z, TargetFunctions -> {Abs, Arg}] // Simplify   
(* huge symbolic expression ...*)
%//N
(*0.523774 - 0.872962 I*)

Answer 2

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

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