
Image and kernel
C* is the group of all nonzero complex numbers under multiplication
H: C*>C*
z>zz (second z is complex conjugate of complex number z)
This is a homomorphism.
The identity element in (C*,x) is 1, so
Ker H = {z E C* : H (z) = 1}
= {z E C* : zz = 1}
= { z E C* : z = z}
= R+
Because z = z (complex conjugate) for each z E C*, the function H is onto and
Im(H) = C*.
Is this correct?
Also I need to find a group with is isomorphic to Ker (H).

Re: Image and kernel
you can't just SAY H is a homomorphism, you have to offer some evidence.
if , to show H is a homomorphism, you need to show that:
for all .
now (you should finish this....)
one way to write is . it should be clear that H cannot possibly be onto, because any image H(z) is always a positive real number. so Im(H) is certainly NOT C*.
also, ker(H) is NOT {1}, and certainly not R+. what is H(2)? what is H(1), or H(i)? even better, what is H(√2/2 + i√2/2)? what does it mean to say ?
the kernel of H has an easy to remember shape.

Re: Image and kernel
Thanks
Would it
H (zw) H (zw)
I still dont understand what the kernel would be or the image.

Re: Image and kernel
since , any element of the image is real and positive (that's a big hint).
ker(H) is the preimage of the identity in C*, which is 1. what complex numbers satisfy ?

Re: Image and kernel
Thanks for your help.
So Im (H) would be the set R squared.
Would ker (H) be i(squared).
By the way how do you go advanced?

Re: Image and kernel
use the tags "[ tex]" and "[ /tex]" (without the quotes or the spaces).
Im(H) is not "R squared" it is the set of all nonzero real squares. there is a much simpler description of this set: the positive reals.
z is often called the modulus of z, it is the same thing as the (euclidean) distance of a point in the complex plane from the origin.
if z = x+iy, then z^2 = x^2 + y^2. what do you call the set of points for which x^2 + y^2 = 1?