HELP!!! Find all fifth roots of unity...

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HELP!!! Find all fifth roots of unity...

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Let's do it the hard way. We want to solve the equation x5−1=(x−1)(x4+x3+x2+x+1)=0. Then we are interested in solving x4+x3+x2+x+1=0. Note the symmetry: if r is a root, 1/r is also a root. Why? Thereafter, we have two ways out. First, we divide everything by x2 to get x2+x+1+1x+1x2=(x+1x)+(x+1x)2−1=u2+u−1=0. What did we do? We noted (x+x−1)2−2=x2+x−2 and substituted xu2+u=1+x−1=u. The solutions for u2+u=1 are −φ and φ−1, φ being the golden ratio. We now have two equations: x+1x=−φ⟹x2+φx+1=0⟹x=±φ−3−−−−−√−φ2 x+1x=φ−1⟹x2+(1−φ)x+1=0⟹x=±−φ−2−−−−−−√+φ−12. You can now manipulate it to get a a+bi form. Alternatively, you can multiply (x−r)(x−r−1), divide x4+x3+x2+x+1 by it and discover which r makes remainder zero. This would be somewhat long, so I won't explictly do it, but here is the idea. shareimprove this answer
The \(n\)th roots of any complex number \(z=re^{it}\) are \(z^{1/n}=r^{1/n}e^{i(t+2\pi k)/n}\) where \(k=0,1,2,\ldots,n-1\). For instance, we know that \(1=e^{i0}\), so \(r=1\) and \(t=0\). This is the case when \(k=0\). One fifth root (fix \(k=1\)) will be \[1^{1/5}=1^{1/5}e^{i(0+2\pi)/5}=e^{i(2\pi/5)}\] Continue the pattern.

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