Hello, im a bit stuck!
Let z = cis theta
Show that a. Modz = 1
and b. 1/z = cis(-theta)
Thanks for your help!
a) You are given:
z=cos(theta)+i.sin(theta)
so:
|z|=sqrt[cos^2(theta)+sin^2(theta)]
but cos^2(theta)+sin^2(theta)=1, so |z|=1.
b) 1/z=1/[cos(theta)+i.sin(theta)]
now multiply top and bottom on the right by cos(theta)-i.sin(theta) to get:
1/z=(cos(theta)-i.sin(theta)/[(cos(theta)+i.sin(theta))(cos(theta)-i.sin(theta))]
.....=(cos(theta)-i.sin(theta)/[cos^2(theta)+sin^2(theta))]
.....=cos(theta)-i.sin(theta)
Now cos(theta)=cos(-theta), and sin(theta)=-sin(-theta), so putting these into the last equation above gives:
1/z=cos(-theta)+i.sin(-theta)
as required.
RonL