Finding the Kernel of a Complex Multiplicative Function

[aliciaislol]

Homework Statement

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.

Homework 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

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?

 

[HallsofIvy]

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.). 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.

 

[aliciaislol]

Thank you, I figured out where it went wrong it was an nth root not power. Thnx again