Tleko (tleko@aol.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.i.sin@ , which gets nearer to being correct ... ;-)
