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Finding the Kernel of f

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider C^x, the multiplicative group of nonzero complex numbers, and let f:C^x --> C^x be defined by f(x)=x^4. Find ker f.


    2. Relevant equations
    C - complex numbers
    e^i2xpi = cos theta + isin theta element oof C
    R - reals
    Z- integers
    where R/Z
    This is the equation we got in class:
    ker f= {x element of R : f(x) =1} = {x element of R: e^(i2xpi)=1} = {x element of R: cos(2xpi) + isin(2xpi) =1} = Z


    3. The attempt at a solution
    Based on the above info:
    ker f= {x element of C^x : f(x) =1} = {x element of C^x : (e^(i2xpi))^4 =1} = {x element of C^x : e^(i8xpi) =1} = {x element of C^x: cos(8xpi) + isin(8xpi) =1}
    Am I doing this right? Is there an easier way?
     
  2. jcsd
  3. Oct 31, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The kernel of f is, by definition, the set of all x such that f(x)= x^4= 1 (I started to write "= 0" but since this group is multiplicative, it is the multiplicative identity, of course.:blushing:). Looks to me like the kernel consists fo the fourth roots of 1.

    Fourth roots not fourth powers. Your "8xpi" (the "x" there just means multiplication, right. It is not a variable. Better to use just "8pi".) seems to be going the wrong way.
     
  4. Oct 31, 2009 #3
    Thank you, I figured out where it went wrong it was an nth root not power. Thnx again:smile:
     
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