# Thread: kernal and image with complex numbers

1. ## kernal and image with complex numbers

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.

2. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
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
Do you mean C ---> C? It is not a homomorphism C* ---> C*.

Originally Posted by fireychariot
Just stuck on the next bit of finding the image and kernel of the groups.
For the first one, can you solve the equation z + 3iz = 0? For the second one, represent z as $\displaystyle re^{i\varphi}$ and solve $\displaystyle z/\bar{z}=1$.

3. ## Re: kernal and image with complex numbers

Originally Posted by emakarov
Do you mean C ---> C? It is not a homomorphism C* ---> C*.

For the first one, can you solve the equation z + 3iz = 0? For the second one, represent z as $\displaystyle re^{i\varphi}$ and solve $\displaystyle z/\bar{z}=1$.
Yes sorry C --> C for the first one

why do I represent z as $\displaystyle re^{i\varphi}$ ?

4. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
why do I represent z as $\displaystyle re^{i\varphi}$ ?
Because every complex number can be represented in polar form and it may simplify the solution.

5. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
C* is the group of non sero complex numbers under multiplication
iv) C* --> C*
z --> z / (the complex conjugate) z
Just stuck on the next bit of finding the image and kernel of the groups.
Recall that $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{{{\left| z \right|}^2}}}$. Thus $\displaystyle \frac{z}{{\overline z }} = \frac{{{z^2}}}{{{{\left| z \right|}^2}}}$

6. ## Re: kernal and image with complex numbers

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?

7. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
So from what I am trying to understand is that the Image is what maps z to z +3iz?
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?

Originally Posted by fireychariot
The kernel is what maps z to the identity, am I right there?
No, what maps is the map. The kernel is what is mapped to identity, and the identity is 0 here.

Originally Posted by fireychariot
How do you know what the identity maybe?
Sorry, "what the identity maybe" is an incomplete sentence; I don't know what it means.

8. ## Re: kernal and image with complex numbers

Sorry it was meant to read how do we work out the identity? You say it is 0 but how did you know that?

9. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
how do we work out the identity? You say it is 0 but how did you know that?
Because 0 is the identity in the group of all complex numbers under addition. It satisfies the group axiom about the identity, and group identity is unique.

10. ## Re: kernal and image with complex numbers

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.

11. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
Have I written that out correct?
Yes, except I am not sure how z + 3iz became z - 4iz.

12. ## Re: kernal and image with complex numbers

ooops wrong question I was looking at. But same applies I take it?

is the Ker = {z in C : z+3iz = 0} = {0}?

13. ## Re: kernal and image with complex numbers

Originally Posted by fireychariot
ooops wrong question I was looking at. But same applies I take it?

is the Ker = {z in C : z+3iz = 0} = {0}?
Yes.

14. ## Re: kernal and image with complex numbers

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??

15. ## Re: kernal and image with complex numbers

What do you think about suggestions in posts #2 and #5? Also, it may be a little easier to find the kernel.

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