Re: " for z>0
- To: mathgroup at smc.vnet.net
- Subject: [mg119678] Re: " for z>0
- From: Vivek Joshi <vivekjjoshi29 at gmail.com>
- Date: Fri, 17 Jun 2011 00:08:52 -0400 (EDT)
- References: <201106160800.EAA13210@smc.vnet.net>
You can use ComplexExpand[...], But is this what you are looking for ? In[17]:= Simplify[Conjugate[Sqrt[-(t Cos[2 kx] Cos[ky])+Cos[kx]+M]],-(t Cos[2 kx] Cos[ky])+Cos[kx]+M>0]//ComplexExpand Out[17]= ((M+Cos[kx]-t Cos[2 kx] Cos[ky])^2)^(1/4) Cos[1/2 Arg[M+Cos[kx]-t Cos[2 kx] Cos[ky]]]-I ((M+Cos[kx]-t Cos[2 kx] Cos[ky])^2)^(1/4) Sin[1/2 Arg[M+Cos[kx]-t Cos[2 kx] Cos[ky]]] Vivek J. Joshi On Thu, Jun 16, 2011 at 4:00 AM, liblenovo <liblenovo at gmail.com> wrote: > In[86]:=Simplify[Conjugate[Sqrt[-(t Cos[2 kx] Cos[ky]) + Cos[kx] + > M] ], -(t Cos[2 kx] Cos[ky]) + Cos[kx] + M > 0] > > Out[86]= Sqrt[-t cos(2 kx) cos(ky)+cos(kx)+M]^\[Conjugate] > > I don't want that ^\[Conjugate], why it is there? > >
- References:
- Simplify expression like " Conjugate[Sqrt[z]] " for z>0
- From: liblenovo <liblenovo@gmail.com>
- Simplify expression like " Conjugate[Sqrt[z]] " for z>0