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Math Help - kernal and image with complex numbers

  1. #16
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    Re: kernal and image with complex numbers

    For the kernel I get so far

    Ker = { z in C* : z / (the complex conjugate) z = 1}

    = { z in C* : z = (the complex conjugate) z }

    Then get stuck. Sorry I have tried looking at #2 and #5 to help but just confueses me a little more.
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  2. #17
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    Re: kernal and image with complex numbers

    Let's use Plato's hint. We need to find all z such that \frac{z^2}{|z|^2}=1. This equality implies that z^2 is a real number, so either z\in\mathbb{R} or z = ir for some r\in\mathbb{R}. In the second case, |z| = r and z^2=-r^2, so \frac{z^2}{|z|^2}=-1. What about the first case?

    If we use the polar form, z=re^{i\varphi} for some r, \varphi\in\mathbb{R}. Then \bar{z}=re^{-i\varphi}, so \frac{z}{\bar{z}}=? What restriction does it put on \varphi?
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  3. #18
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    Re: kernal and image with complex numbers

    suppose f(z) = z/z* is our homomorphism.

    if z is in ker(f), then z/z* = 1, so

    z = z*
    z-z* = 0
    (1/(2i))(z-z*) = Im(z) = 0.

    thus all elements of the kernel are real.

    if however, z is real, then z = z*, so z/z* = 1, thus z is in ker(f).

    so ker(f) = R

    note that |f(z)| = |z/z*| = |z|/|z*| = |z|/|z| = 1. thus im(f) is a subset of the unit circle.

    on the other hand, suppose that w is any complex number on the unit circle (that is: |w| = 1).

    there exists at least one complex number z with z2 = w, and from |z|2 = |z2| = |w| = 1, we see that |z| = 1, as well.

    so f(z) = z/z* = (z/z*)(1) = (z/z*)(z/z) = z2/(zz*) = z2/|z|2 = z2 = w, which means that im(f) is the entire unit circle.
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