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Math Help - kernal and image with complex numbers

  1. #1
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    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.
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  2. #2
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    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*.

    Quote Originally Posted by fireychariot View Post
    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 re^{i\varphi} and solve z/\bar{z}=1.
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    Re: kernal and image with complex numbers

    Quote Originally Posted by emakarov View Post
    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 re^{i\varphi} and solve z/\bar{z}=1.
    Yes sorry C --> C for the first one

    why do I represent z as re^{i\varphi} ?
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    why do I represent z as re^{i\varphi} ?
    Because every complex number can be represented in polar form and it may simplify the solution.
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    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 \frac{1}{z} = \frac{{\overline z }}{{{{\left| z \right|}^2}}}. Thus \frac{z}{{\overline z }} = \frac{{{z^2}}}{{{{\left| z \right|}^2}}}
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    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?
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    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?

    Quote Originally Posted by fireychariot View Post
    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.

    Quote Originally Posted by fireychariot View Post
    How do you know what the identity maybe?
    Sorry, "what the identity maybe" is an incomplete sentence; I don't know what it means.
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  8. #8
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    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?
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    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.
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    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.
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  11. #11
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    Re: kernal and image with complex numbers

    Quote Originally Posted by fireychariot View Post
    Have I written that out correct?
    Yes, except I am not sure how z + 3iz became z - 4iz.
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  12. #12
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    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}?
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    Re: kernal and image with complex numbers

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

    is the Ker = {z in C : z+3iz = 0} = {0}?
    Yes.
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  14. #14
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    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??
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  15. #15
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    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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