Hello, im a bit stuck!

Let z = cis theta

Show that a. Modz = 1

and b. 1/z = cis(-theta)

Thanks for your help!

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- Feb 1st 2007, 01:24 AMclassicstringsComplex Numbers
Hello, im a bit stuck!

Let z = cis theta

Show that a. Modz = 1

and b. 1/z = cis(-theta)

Thanks for your help! - Feb 1st 2007, 04:02 AMCaptainBlack
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 - Feb 2nd 2007, 08:10 PMclassicstrings
Thanks for your help CB!