Go backward to RelationshipGo up to TopGo forward to Proof (Continued) |

Take arbitraryforallxinC',yinC': cartesian(x*_{C'}y) = cartesian(x) *_{C}cartesian(y).

(*) holds because of the definition "shift" and for every

cartesian( x*_{C'}y)= cartesian( x_{0}y_{0}, shift(x_{1}+y_{1}))= x_{0}y_{0}cos(shift(x_{1}+y_{1})) + (x_{0}y_{0}sin(shift(x_{1}+y_{1})))i= (*) x_{0}y_{0}cos(x_{1}+y_{1}) + (x_{0}y_{0}sin(x_{1}+y_{1}))i.

sin( x+2 pi )= sin( x),cos( x+2 pi )= cos( x)

Author: Wolfgang Schreiner

Last Modification: December 7, 1999