Question

show that the roots of the equation z^4-1=0 form a cyclic group of order 4.

Answer 1

$Z^4-1=0\\ Z^4=1\\ Z^4=e^{2\pi n}\quad n=0,1,2,3\\ Z=e^{\frac{2\pi n}{4}}\quad n=0,1,2,3\\$

therefore,

$Z_1=1 \\ Z_2=e^{\frac{\pi}{2}}=j\\ Z_3=e^{\pi}=-1\\ Z_4=e^{\frac{3\pi}{2}}=-j\\$

$A=\{1,j,-1,-j\}=\{z^0,z^1,z^2,z^3\},\quad z=j.\\ \therefore A=<z>=\{z^k|k=0,1,2,3\}$

Therefore set A is a cyclic group of order 4.