1

If I input: 

Solve[z^4 == 2 (Cos[2 Pi/3] + I Sin[2 Pi/3])]

Mathematica returns: 

{{z -> -(-1 + I Sqrt[3])^(1/4)}, {z -> -I (-1 + I Sqrt[3])^(1/4)}, 
 {z -> I (-1 + I Sqrt[3])^(1/4)}, {z -> (-1 + I Sqrt[3])^(1/4)}}.

I want to see these complex numbers in their polar form. For example, one of the roots in the list is 2^(1/4)*(Cos[Pi/6] + I Sin[Pi/6])

m_goldberg
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Geoffrey Critzer
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2 Answers2

2

The roots as Cartesian complex numbers.

roots = Solve[z^4 == 2 (Cos[2 Pi/3] + I Sin[2 Pi/3])][[All, 1, 2]]

{-(-1 + I Sqrt[3])^(1/4), -I (-1 + I Sqrt[3])^(1/4), 
 I (-1 + I Sqrt[3])^(1/4), (-1 + I Sqrt[3])^(1/4)}

The roots converted to polar form.

 polarRoots = AbsArg /@ roots

{{2^(1/4), Arg[-(-1 + I*Sqrt[3])^(1/4)]}, 
 {2^(1/4), Arg[(-I)*(-1 + I*Sqrt[3])^(1/4)]}, 
 {2^(1/4), Arg[I*(-1 + I*Sqrt[3])^(1/4)]}, 
 {2^(1/4), Pi/6}}

These values may not look the same as the representation you were expecting, but to Mathematica they really are the same, as can be seen by plotting polarRoots.

ListPolarPlot[Reverse /@ polarRoots,
  PolarAxes -> True,
  PlotStyle -> {Red, AbsolutePointSize[15]}]

plot

m_goldberg
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  • Thank You! I am still using version 9. polarRoots = AbsArg /@ roots didn't work but I adapted with: polarRoots = Map[{Abs[#], Arg[#]} &, roots] – Geoffrey Critzer Dec 13 '15 at 12:47
1

Following Nasser's suggestion...

polarForm = Expand[# /. z_?NumericQ :> Abs[z] Exp[I Arg[z]]] &;

Then 

 Solve[z^4 == polarForm[ (Cos[2 Pi/3] + I Sin[2 Pi/3])]]

(* {{z->-Power[-1, (6)^-1]},{z->Power[-1, (6)^-1]},{z->-(-1)^(2/3)},{z->(-1)^(2/3)}} *)

chris
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  • But I want to see ALL the roots in their polar form. I tried this: Map[Abs[z /. # ] Exp[I Arg[z /. #]] &, Solve[z^4 == 2 (Cos[2 Pi/3] + I Sin[2 Pi/3])]]. I got only one root in its polar form. The problem seems to be that Arg[z] doesn't seem to work on some of the roots? – Geoffrey Critzer Dec 12 '15 at 21:44