Example 1.2.14 (b): Multiplying geometrically 
Draw the vectors z^{n}, n=1, 2, ... 8 for
z=cis(/4) and
z=0.85 cis(/8)
Context

z^{n} means to multiply z with itself, and
we know how to interpret multiplication geometrically. Thus, with
z=cis(/4) we have:
 z=cis(/4)
 z^{2}=z*z=
cis(/4)*cis(/4) =
cis(/4+/4)=
cis(/2)
 z^{3}=z*z^{2}=
cis(/4)*cis(/2) =
cis(/4+/2)=
cis(3/4)
 z^{4}=z*z^{3}=
cis(/4)*cis(3/4) =
cis(/4+3/4)=
cis()
 z^{5}=z*z^{4}=
cis(/4)*cis() =
cis(/4+)=
cis(5/4)
 z^{6}=z*z^{5}=
cis(/4)*cis(5/4) =
cis(/4+5/4)=
cis(3/2)
 z^{7}=z*z^{6}=
cis(/4)*cis(3/2) =
cis(/4+3/2)=
cis(7/4)
 z^{8}=z*z^{7}=
cis(/4)*cis(7/4) =
cis(/4+7/4)=
cis(2)=1


The vector z=1/2 cis(/8) has
similar powers, adjusting for the fact that its length is not one:
 z=0.85 cis(/4)
 z^{2}=z*z=
0.85 cis(/4) * 0.85 cis(/4) =
0.85^{2} cis(/4+/4)=
0.85^{2} cis(/2)
 z^{3}=z*z^{2}=
0.85 cis(/4) * 0.85^{2} cis(/2) =
0.85^{3} cis(/4+/2)=
0.85^{3} cis(3/4)
 z^{4}=z*z^{3}=
0.85 cis(/4) * 0.85^{3} cis(3/4) =
0.85^{4} cis(/4+3/4)=
0.85^{4} cis()
 z^{5}=z*z^{4}=
0.85 cis(/4) * 0.85^{4} cis() =
0.85^{5} cis(/4+)=
0.85^{5} cis(5/4)
 z^{6}=z*z^{5}=
0.85 cis(/4) * 0.85^{5} cis(5/4) =
0.85^{6} cis(/4+5/4)=
0.85^{6} cis(3/2)
 z^{7}=z*z^{6}=
0.85 cis(/4) * 0.85^{6} cis(3/2) =
0.85^{7} cis(/4+3/2)=
0.85^{7} cis(7/4)
 z^{8}=z*z^{7}=
0.85 cis(/4) * 0.85^{7} cis(7/4) =
0.85^{8} cis(/4+7/4)=
0.85^{8}0.85^{2} cis(2)=1


Interactive Complex Analysis, ver. 1.0.0
(c) 20062007, Bert G. Wachsmuth
Page last modified: May 29, 2007