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Re: TrigToExp of ArcTan function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38565] Re: [mg38547] TrigToExp of ArcTan function
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 27 Dec 2002 02:14:54 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

On Wednesday, December 25, 2002, at 05:12 PM, Blimbaum Jerry DLPC wrote:

>
>
> Why does   Cos[ArcTan[x,y]] + I Sin[ArcTan[x,y]]//TrigToExp   not give 
> an
> output   of
>
>
> Exp[I ArcTan[x,y]]   ??
>
>
> Instead the output  is    x/Sqrt[x^2+y^2] +   I y/Sqrt[x^2 + 
> y^2]........
>
>
> thanks     jerry blimbaum   NSWC  panama city, fl
>
>
>

The reason is simply that TrigToExp acts on each term separately thus:

In[1]:=
TrigToExp[Cos[ArcTan[x, y]]]

Out[1]=
x/(2*Sqrt[x^2 + y^2]) + (I*y)/(2*Sqrt[x^2 + y^2]) +
   Sqrt[x^2 + y^2]/(2*(x + I*y))

In[2]:=
TrigToExp[I*Sin[ArcTan[x, y]]]

Out[2]=
x/(2*Sqrt[x^2 + y^2]) + (I*y)/(2*Sqrt[x^2 + y^2]) -
   Sqrt[x^2 + y^2]/(2*(x + I*y))

In[3]:=
%% + %

Out[3]=
x/Sqrt[x^2 + y^2] + (I*y)/Sqrt[x^2 + y^2]


If you are willing to assume that x and y are real then one way to get 
an output involving E  is:

In[4]:=
ComplexExpand[Cos[ArcTan[x, y]] + I*Sin[ArcTan[x, y]],
   TargetFunctions -> {Arg}]

Out[4]=
Cos[Arg[x + I*y]] + I*Sin[Arg[x + I*y]]

In[5]:=
TrigToExp[%]

Out[5]=
E^(I*Arg[x + I*y])

without that assumption you can get a more complicated answer:

In[6]:=
ComplexExpand[Cos[ArcTan[x, y]] + I*Sin[ArcTan[x, y]],
   {x, y}, TargetFunctions -> {Arg, Abs}]

Out[6]=
(Abs[x + I*y]*Cos[Arg[(x + I*y)/Sqrt[x^2 + y^2]]])/
    Sqrt[Abs[x^2 + y^2]] +
   (I*Abs[x + I*y]*Sin[Arg[(x + I*y)/Sqrt[x^2 + y^2]]])/
    Sqrt[Abs[x^2 + y^2]]

In[7]:=
TrigToExp[%]

Out[7]=
(E^(I*Arg[(x + I*y)/Sqrt[x^2 + y^2]])*Abs[x + I*y])/
   Sqrt[Abs[x^2 + y^2]]


Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/


Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/



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