A Nice Quintic | Video Response to @blackpenredpen

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If you write the equation as x^5 = e^(2 i π ) then five solutions are x= 1, x=e^(2 i π k/5), k= 1, 2, 3, 4 . They form a regular pentagon on the unit circle in the complex plane. Analogously, you get an n - gon as solutions of x^n = 1.

renesperb
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Looks like you forgot the minus sign in the second solution for t....

lancediduck
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I just expressed 1 in polar form from the start to get the rest of the solutions using Euler's number.

scottleung
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Paul, I was with you, right until 8:47, when you dropped a negative sign.

dixonblog
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This is a nice way to get sin and cos values for 72, 144 ... Degrees
Forgotten all about palindromes and i was waiting for you to match coefficients for quadratics :)

dan-florinchereches
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That's simple:-
Equation with nth degree has n number of roots.
We know,
e^2nπi = 1
Where n belong to integer and i is an imaginary number.
x = e^(2nπi/5)
Let n = -1, -2, 0, 1, 2
Then x1 = cos(0.4π)-isin(0.4π)
x2 = conjugate of x1
x3 = 1
x4 = cos(0.8π)+isin(0.8π)
x5 = conjugate of x4

Thank you.

gauravroy
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Assume a solution is written in the complex plane as r e (i Theta) ; you immediately get that r = 1 to satisfy the solution. So you are left looking for theta. roots 2 k pi / 5 are straightforward after. Roots of unity are so, so, so useful.
now if you really really want to write down the algebraic value sin and cos of 2 pi /5 .. then go ahead, but I feel that students will remember more readily the roots of unity e (2k i pi/5) rather than a monstruous algebraic expression.

sugamonogaijin
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Sums of coefficients of odd and even powers are equal. That starts us with one solution, and we need to look for 4 more.

neuralwarp
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Can you figure out the SEVENTH roots of unity? (I dare ya!)

Blaqjaqshellaq
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Complexify 1 and solve. I almost bet my fortune that you would do it that way and call it "method 2"

trojanleo
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x^5+/-x^4+/-x^3+/-x^2+/-x-1=0, (x-1)(x^4+x^3+x^2+x+1)=0, x^4+x^3+x^2+x+1=0,
1 -1
1 -1
1 -1
1 -1
1 -1

prollysine
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