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

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."

• You could simplify $\cos\left(\frac\pi2\right),$ etc. Commented Jan 26, 2020 at 18:34
• converting eulers to polar Commented Jan 26, 2020 at 18:36
• directly came from an earlier question math.stackexchange.com/questions/3522808/… Commented Jan 26, 2020 at 18:37
• how? how do I use the unity circle when its imaginary answer Commented Jan 26, 2020 at 18:46