1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook