Question

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

Answer 1

@freckles

Answer 2

@MAli13chineta

Answer 3

@geerky42

Answer 4

@whpalmer4

Answer 5

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

Answer 6

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.