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Math Help - Homomorphism #2

  1. #1
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    Homomorphism #2

    Is this correct anyone;

    C is the group of all complex numbers under addition

    O : C ----> C
    z ----> z + iz

    For all z1,z2 E C

    O (z1+z2) = z1 + z2 + iz1 + iz2 = z1+iz1 + z2+iz2 = O (z1) + O (z2)

    Hence O satisfies the homomorphism property and so is a homomorphism.
    Last edited by mr fantastic; June 21st 2011 at 01:50 PM. Reason: Title.
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  2. #2
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    Re: Homomorphism #2

    Quote Originally Posted by Arron View Post
    Is this correct anyone;

    C is the group of all complex numbers under addition

    O : C ----> C
    z ----> z + iz

    For all z1,z2 E C

    O (z1+z2) = z1 + z2 + iz1 + iz2 = z1+iz1 + z2+iz2 = O (z1) + O (z2)

    Hence O satisfies the homomorphism property and so is a homomorphism.


    It's pretty easy to show that multiplication by any complex number is a group homomorphism of \mathbb{C} , and your function is only

    multiplication by 1+i...

    Tonio
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  3. #3
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    Re: Homomorphism #2

    explicitly, let w \in \mathbb{C} be given, and define:

    h_w:\mathbb{C} \rightarrow \mathbb{C} by h_w(z) = zw

    then h_w(z_1+z_2) = (z_1+z_2)w = z_1w+z_2w = h_w(z_1) + h_w(z_2)

    this is what distributivity tells us in a commutative ring R: that multiplication by a fixed element r is an additive endomorphism. a slightly more archaic way of saying this, is that multiplication is compatible with addition.
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  4. #4
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    Re: Homomorphism #2

    I am trying to find the image and kernal of this homomorphism, am I correct in saying

    The identity element in (C, +) is 0, so
    Ker O = {z E C : O (z) =0}
    = (z E C : z+iz = 0}
    = {0}

    Because z + iz = z for each z E C, the function O is onto and
    Im (O) = C.
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  5. #5
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    Re: Homomorphism #2

    saying what you wish to be true, is not what constitutes a "proof" or demonstration.

    in particular, if you claim that ker(O) = {0}, you have to give a reason for it.

    one possible justification is that O(z) = z+iz = z(1+i). so if O(z) = z(1+i) = 0,

    then z(1+i)((1-i)/2)) = 0((1-i)/2) --> z = 0. this shows that ker(O) is a subset of {0},

    and obviously we have 0 is an element of ker(O), so ker(O) = {0}.

    now, your argument that O is onto is just not one at all.

    in the first place, z+ iz DOES NOT EQUAL z (unless z = 0).

    to prove O is onto, given any w in C, you have to FIND some z in C with O(z) = w.

    to do this, you first set w = z + zi = z(1+i), and try to solve for z.

    then z = w/(1+i).

    does this work? well: O(w/(1+i)) = (w/(1+i)) + (w/(1+i))i = (w/(1+i))(1+i) = w((1+i)/(1+i)) = w(1) = w.

    THAT shows O is onto, so that Im(O) = C.
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  6. #6
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    Re: Homomorphism #2

    Hi,

    Regarding your answer to the above homomorphism.

    Could you not have divided each side by (1 + i), instead of multiplying by (the conjugate/2)?

    Thank you,

    GaryOg.
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  7. #7
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    Re: Homomorphism #2

    one could do many things.

    for example, one could use the fact that since C is a field, it is a fortiori, an integral domain, so:

    z(1+i) = 0, implies one of z or 1+i is 0. since 1+i is not 0, z must be.

    dividing by 1+i is the same as multiplying by 1/(1+i).

    since (1+i)((1-i)/2)) = (1/2)(1+i)(1-i) = (1/2)(12 - i2) = (1/2)(1 - (-1)) = (1/2)(1 + 1) = (1/2)(2) = 1,

    we see that 1/(1+i) *IS* (1-i)/2. in other words, "dividing by 1+i" is exactly what i did.
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