In article <firstname.lastname@example.org> Hunter James <email@example.com> wrote:
: I see what you mean. And with that observation, we have the jingo: : : y is an analytic function of both z and z*, if, and only if, : it is a "function" of neither. : :>tleko wrote: :> Indeed we have z=z* . :> :> We write z=r.exp(i.@) :> :> z*=r.exp(i.(-@)) :> :> to obtain z-z*=r.((cos@+i.sin@)-(cos(-@)+i.sin(-@-pi))) = 0 . :>) In article <199610070856.KAA0189@maloche.neuroinformatik. :>) ruhr-uni-bochum> David Kastrup wrote: :>) This above all proves that you are not wanting to make any sensible :>) claims but are just out to "prove" nonsense by mixing your geometric :>) functions up. The "-pi" in the above is absolute nonsense and not :>) justifiable at all. If you cross out that junk, you arrive at :>) z-z* = 2 r i sin@ :>) :>) Which is, pretty obvious for all bu the hopefully dense, the 2i Im(z) :>) which you could have arrived at without employing polar coordinates :>)@ In article <firstname.lastname@example.org> Andreas Leitgeb wrote: :>)@ Tleko (email@example.com) wrote: :>)@> Indeed we have z=z* . :>)@nope :>)@> We write z=r.exp(i.@) :>)@> z*=r.exp(i.(-@)) :>)@yup :>)@> to obtain z-z*=r.((cos@+i.sin@)-(cos(-@)+i.sin(-@-pi))) = 0 . :>)@where does _this_ come from ?? ---------------------^^^ :>)@how about: :>)@z-z*=r.((cos@+i.sin@)-(cos(-@)+i.sin(-@))) = :>)@ r. (cos@+i.sin@)-(cos(@) -i.sin(@))) = :>)@ r. ( i.sin@ - ( - i.sin@ ) = 2.r.i.sin@ :>)@ which gets nearer to being correct ... ;-) :>)@@In the article <firstname.lastname@example.org> Ilias Kastanas wrote: :>)@@Is it possible not to realize this can only hold for y=o, i.e. for real :>)@@z, :>)@@and not in general?
Yes it is. 2.r.i.sin@ = 2.i.y is valid for any real y not only for y=0.