1. ## Image and kernel

C* is the group of all non-zero 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).

2. ## Re: Image and kernel

you can't just SAY H is a homomorphism, you have to offer some evidence.

if $H(z) = z\overline{z}$, to show H is a homomorphism, you need to show that:

for all $z,w \in \mathbb{C}^*, H(zw) = H(z)H(w)$.

now $H(zw) = (zw)(\overline{zw}) = zw(\overline{z})(\overline{w}) = \dots ?$ (you should finish this....)

one way to write $z\overline{z}$ is $|z|^2$. 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 $|z|^2 = 1$?

the kernel of H has an easy to remember shape.

3. ## Re: Image and kernel

Thanks

Would it

H (zw) H (zw)

I still dont understand what the kernel would be or the image.

4. ## Re: Image and kernel

since $H(z) = |z|^2$, any element of the image is real and positive (that's a big hint).

ker(H) is the pre-image of the identity in C*, which is 1. what complex numbers satisfy $|z|^2 = 1$?

5. ## Re: Image and kernel

So Im (H) would be the set R squared.

Would ker (H) be i(squared).

By the way how do you go advanced?

6. ## 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 non-zero 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?