find the $4th$ and $5th$ roots of unity

Question

I am looking for solution verification. This is my first time doing roots of unity

$$z^4=1$$

$k=0, \space \space z=e^0 = \cos(0)+i\sin(0)=1$

$k=1,\space \space z = e^{\frac{2 \pi}{4}i}=e^{\frac{\pi}{2}}=\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$

$k = 2 \space \space z = e^{\pi i} = \cos(\pi)+ i\sin(\pi)$

$k = 3 \space \space z = e^{\frac{3 \pi}{2}i} = \cos(\frac{3 \pi}{2}+i\sin(\frac{3 \pi}{2}) = -1$

$$z^5=1$$

$-1$ is excluded as a possible answer because the power is an odd power.

$k = 0 \space \space z = e^{0}= \cos(0)+i\sin(0) = 1$

$k = 1 \space \space z = e^{\frac{2 \pi}{5}i}= \cos(\frac{2 \pi}{5})$

$k = 2 \space \space z = e^{\frac{4 \pi}{5}i} = \cos(\frac{4 \pi}{5})+i\sin(\frac{4 \pi}{5})$

$k = 3 \space \space z = e^{\frac{6 \pi}{5}i}= \cos(\frac{6 \pi}{5})+i\sin(\frac{6 \pi}{5})$

$k = 4 \space \space z = e^{\frac{8 \pi}{5}i} = \cos(\frac{8 \pi}{5} + i\sin(\frac{8 \pi}{5})$

"with some slightly more advanced mathematical knowledge we have derived a simple formula to find all the n-th roots of unity, for any n. The formula we came up with last time is:

The n, all distinct, n-th roots of unity are cos (2kpi/n) + i sin (2kpi/n), k= 0, 1, ... , n-1."

Answer 1

Apart from some missing brackets on the cos terms, all seems correct apart from :

• When k = 3 for the first part, cos(3pi/2) + isin(3pi/2) = -i rather than -1

you can simplify for the first part:

for k = 1 , = i

for k = 2 , = -1

The solution is symmetrical with each root 90 degrees from the other, forming a square in the complex plane