Convert a complex number into the exponent form

Questions
1

See my following code snippet:

In[344]:= (*
https://mathematica.stackexchange.com/questions/16414/how-can-i-convert-a-complex-number-into-an-exponent-form
*)
polarForm = Expand[# /. z_?NumericQ :> Abs[z] Exp[I Arg[z]]] &;
polarForm1 = ComplexExpand[# /. z_?NumericQ :> Abs[z] Exp[I Arg[z]], TargetFunctions -> {Re, Im}] &;
x1=Eigenvectors[gen1]
polarForm/@#&/@x1
polarForm1/@#&/@x1

Out[346]= {{1, 1, 1}, {1/2 (-1 - I Sqrt[3]), 1/2 (-1 + I Sqrt[3]), 
  1}, {1/2 (-1 + I Sqrt[3]), 1/2 (-1 - I Sqrt[3]), 1}}

Out[347]= {{1, 1, 1}, {E^(-((2 I \[Pi])/3)), E^((2 I \[Pi])/3), 
  1}, {E^((2 I \[Pi])/3), E^(-((2 I \[Pi])/3)), 1}}

Out[348]= {{1, 1, 1}, {-(1/2) - (I Sqrt[3])/2, -(1/2) + (I Sqrt[3])/2,
   1}, {-(1/2) + (I Sqrt[3])/2, -(1/2) - (I Sqrt[3])/2, 1}}

Why doesn’t the second method work?

Regards,
Zhao

2

ComplexExpand always separates real and imaginary parts. This is not what one wants in a polar form. So it cannot be used to create that form, or at least cannot be used in this particular manner.

3

Based on your title,

Convert a complex number into the exponent form.

if you’re just converting numbers, then you don’t need to expand. You could just do this:

ToPolar[z_] := Abs[z] Exp[I Arg[z]]

Abs and Arg are listable, so you can apply it like this:

ToPolar[x1]

If this isn’t what you’re looking for, then provide more context. If your question is specifically about the behavior of ComplexExpand, then be explicit about your question (but Daniel’s answer is likely what you’re looking for).

Throwing some code at us and asking “why doesn’t it work” imposes a burden on us to infer your objective and what you think is wrong.

4

Thank you, Eric,

This is exactly what I’m looking for.