Answer
a. $\pm1,\pm i$
b. $1,\frac{-1\pm \sqrt 3i}{2}$
c. 6 roots.
d. n roots.
Work Step by Step
For $x^4-1=0$, there are 4 roots $\pm1,\pm i$
For $x^3-1=0$, there are 3 roots $1,\frac{-1\pm \sqrt 3i}{2}$
For $x^6-1=0$, $(x^3-1)(x^3+1)=0$, let $x^3-1=0$ and $x^3+1=0$, we can obtain 6 roots.
For $x^n-1=0$, it is expected that there will be n number of roots.
Extra information:
$x^n=cos(0)+i\cdot sin(0)$, the $nth$ root of 1 can be obtained as
$x_n=cos\frac{2k\pi}{n}+i\cdot sin\frac{2k\pi}{n}, k=0,1,...n-1$
