Roots of Unity - is this correct?

[Kaldanis]

I'm going over some things I didn't do too well on in my latest Algebra test.
One question was: List all of the roots of $x^{8}\:-1\:=0$, and write them in the form $a+bi$.

So I knew I had to list all the 8th roots of unity. In other places in the test they used the notation $e^{iθ}$ and this was a huge problem for me, I didn't know that $r\ast e^{iθ}$= $r*(cos(θ)\:+isin(θ))$.

What I'm looking for: $e^{i*2\pi*\frac{k}{n}}$, or in this case, $e^{i*2\pi*\frac{k}{8}}$, where k = 0, 1, 2, 3, 4, 5, 6, 7. (Could also write this as $cos(\frac{2\pi*k}{8})\:+isin(\frac{2\pi*k}{8})$ ).

For the 1^st root: $cos(\frac{2\pi*0}{8})\:+isin(\frac{2\pi*0}{8})\:=cos(0)\:+isin(0)$ = $1+0i\:=1$

For the 2^nd root: $cos(\frac{2\pi*1}{8})\:+isin(\frac{2\pi*1}{8})\:=cos(\frac{\pi}{4})\:+isin(\frac{\pi}{4})\:=\frac{ \sqrt{2}}{2}+\frac{ \sqrt{2}}{2}i$

For the 3^rd root: $cos(\frac{2\pi*2}{8})\:+isin(\frac{2\pi*2}{8})\:=cos(\frac{\pi}{2})\:+isin(\frac{\pi}{2})\:=0+1i\:=i$

For the 4^th root: $cos(\frac{2\pi*3}{8})\:+isin(\frac{2\pi*3}{8})\:=cos(\frac{3\pi}{4})\:+isin(\frac{3\pi}{4})\:=-\frac{ \sqrt{2}}{2}+\frac{ \sqrt{2}}{2}i$

5th root: -1
6th root: $-\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i$
7th root: -i
8th root:$\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i$So the correct answer to this question is $±1,\:\:±i,\:\:and\:\:±\frac{ \sqrt{2}}{2}±\frac{ \sqrt{2}}{2}i$Is this right? And if I just apply the same method to other similar problems I should be fine?

 

[ehild]

Is this right? And if I just apply the same method to other similar problems I should be fine?

Congratulation, you did a good job! Your derivation is rigth, and you can apply this method for all roots of unity.
If you need to find the roots of an equation zⁿ-r e^iψ=0 do the following:

$$z=(r\exp(i\psi))^{1/n}(e^{i 2\pi k/n})=r^{1/n} e^{i(\psi /n+2\pi k/n)}=r^{1/n} \left( \cos(\psi /n+2\pi k/n)+i \sin(\psi /n+2\pi k/n)\right)$$

ehild