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/