For the kernel I get so far
Ker = { z in C* : z / (the complex conjugate) z = 1}
= { z in C* : z = (the complex conjugate) z }
Then get stuck. Sorry I have tried looking at #2 and #5 to help but just confueses me a little more.
For the kernel I get so far
Ker = { z in C* : z / (the complex conjugate) z = 1}
= { z in C* : z = (the complex conjugate) z }
Then get stuck. Sorry I have tried looking at #2 and #5 to help but just confueses me a little more.
Let's use Plato's hint. We need to find all z such that . This equality implies that is a real number, so either or z = ir for some . In the second case, |z| = r and , so . What about the first case?
If we use the polar form, for some . Then , so What restriction does it put on ?
suppose f(z) = z/z* is our homomorphism.
if z is in ker(f), then z/z* = 1, so
z = z*
z-z* = 0
(1/(2i))(z-z*) = Im(z) = 0.
thus all elements of the kernel are real.
if however, z is real, then z = z*, so z/z* = 1, thus z is in ker(f).
so ker(f) = R
note that |f(z)| = |z/z*| = |z|/|z*| = |z|/|z| = 1. thus im(f) is a subset of the unit circle.
on the other hand, suppose that w is any complex number on the unit circle (that is: |w| = 1).
there exists at least one complex number z with z^{2} = w, and from |z|^{2} = |z^{2}| = |w| = 1, we see that |z| = 1, as well.
so f(z) = z/z* = (z/z*)(1) = (z/z*)(z/z) = z^{2}/(zz*) = z^{2}/|z|^{2} = z^{2} = w, which means that im(f) is the entire unit circle.