I'm going over some things I didn't do too well on in my latest Algebra test.(adsbygoogle = window.adsbygoogle || []).push({});

One question was: List all of the roots of [itex]x^{8}\:-1\:=0[/itex], and write them in the form [itex]a+bi[/itex].

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

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

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

For the 2^{nd}root: [itex]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[/itex]

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

For the 4^{th}root: [itex]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[/itex]

5th root: -1

6th root: [itex]-\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i[/itex]

7th root: -i

8th root:[itex]\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i[/itex]

So the correct answer to this question is [itex]±1,\:\:±i,\:\:and\:\:±\frac{ \sqrt{2}}{2}±\frac{ \sqrt{2}}{2}i[/itex]

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

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# Homework Help: Roots of Unity - is this correct?

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