C^{*} is the group of non sero complex numbers under multiplication, and C is the group of all complex numbers under addition.
iii) C* ---> C*
z --> z + 3iz
iv) C* --> C*
z --> z / (the complex conjugate) z
I have already proved that these are homomorphism. Just stuck on the next bit of finding the image and kernel of the groups.
Because every complex number can be represented in polar form and it may simplify the solution.
I am sorry but I still don't know how to work each out. Start with the image for the first one
iii) C ---> C
z --> z + 3iz
So from what I am trying to understand is that the Image is what maps z to z +3iz? Am I thinking in the right terms there? The kernel is what maps z to the identity, am I right there? How do you know what the identity maybe?
No, what maps z to z +3iz is the map, the homomorphism. Why don't you look up the definition of image in your textbook or Wikipedia?
No, what maps is the map. The kernel is what is mapped to identity, and the identity is 0 here.
Sorry, "what the identity maybe" is an incomplete sentence; I don't know what it means.
Oh right thanks good point to know. So for the image of
iii) C ---> C
z --> z + 3iz
Im = { w in C : w = z - 4iz, some z in C}
= { w in C : w/(1-4i) in C}
= C
Have I written that out correct? If so I will have a go at the kernel next.
ok back to the next question then,
iv) C* --> C*
z --> z / (the complex conjugate) z
Im = { w in C* : w = z / (the complex conjugate) z, for some z in C*}
However stuck on how to maniplulate this for the next line??