In article <325bb0af.0@news.cranfield.ac.uk> Simon Read <s.read@cranfield.ac.uk> wrote:
tleko@aol.com (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 . ^^^^^ Simon Read <s.read@cranfield.ac.uk> wrote: >> What? Where did that -pi come from? We're just replacing @ with >> -@ >> >> z-z* = r.( (cos@+i.sin@) - (cos(-@)+i.sin(-@)) ) >> >> = r. ( cos@ - cos@ + i.sin@ -i.sin(-@) ) >> >> = r. ( i.sin@ + i.sin@ ) >> >> = 2 i r sin@ >>> In the article <53e100$60r@nuke.csu.net> Ilias Kastanas wrote: >>> Is it possible not to realize this can only hold for y=0 i.e. >>> for real z, and not in general?
Yes it is . 2 i r sin@ = 2 i y is valid for any real y not only for y=0.
