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/