Re: problem of evaluating SQRT
- To: mathgroup at smc.vnet.net
- Subject: [mg28051] Re: problem of evaluating SQRT
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Thu, 29 Mar 2001 03:24:16 -0500 (EST)
- References: <99pd51$leb@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Pek Simplify[Sqrt[z^2]-z, z>=0] 0 Simplify[Sqrt[z^2]-z, -Pi/2< Arg[z] <=Pi/2] 0 Here is what is going on Sqrt assumes that we are working with complex numbers. For real non-negative r it gives the normal non-negative square root of r For non-zero complex numbers z= x+Iy and r = Sqrt[x^2+y^2], the principal argument (angle) of z is the angle a such that z = r (Cos[a]+ I Sin[a]) and -Pi<a<=Pi. Sqrt[z] is the principal square root of z, that is Sqrt[r]( Cos[a/2]+ r I Sin[a/2] ) Now, z^2 = r^2(Cos[2a]+ I Sin[2a]) The question is: what is the principal argument of z^2? (we know that -Pi<a<=Pi). If -Pi/2<a<=Pi/2, and then it is 2a , so Sqrt[z^2] = r (Cos[a]+ I Sin[a]) = z If Pi/2 <a<=Pi then it is 2a -2Pi, so Sqrt[z^2] = r (Cos[a-Pi]+ I Sin[a-Pi]) = -z If -Pi <a<=-Pi/2 then it is 2a +2Pi, so Sqrt[z^2] = r (Cos[a+Pi]+ I Sin[a+Pi]) = -z -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Pek" <phsoh at alum.mit.edu> wrote in message news:99pd51$leb at smc.vnet.net... > Hi, > > We have a question of how sqrt can be evaluated. > > In[1]:= > Sqrt[x^2] > > Out[1]= > (This part is just sqrt[X^2]) > > Below we expect the result to be zero but it isn't. How can we get the > correct answer in this case? > > In[2]:= > Sqrt[x^2] - x > > Out[2]= > (This part is -x + sqrt[x^2] ) > > Will really appreciate your help. Thanks. > > Pek > >