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.
@lancediduck62782 күн бұрын
Looks like you forgot the minus sign in the second solution for t....
@dan-florinchereches48922 күн бұрын
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 :)
@dixonblog2 күн бұрын
Paul, I was with you, right until 8:47, when you dropped a negative sign.
@SyberMath2 күн бұрын
😲😁😄
@scottleung95872 күн бұрын
I just expressed 1 in polar form from the start to get the rest of the solutions using Euler's number.
@SyberMath2 күн бұрын
nice work
@sugamonogaijin14 сағат бұрын
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.
@neuralwarp2 күн бұрын
Sums of coefficients of odd and even powers are equal. That starts us with one solution, and we need to look for 4 more.
@SyberMath2 күн бұрын
are you sure?
@gerryiles39252 күн бұрын
@@SyberMath Given he said "we need to look for 4 more", I think he meant the coefficients of the "original" x^5 = 1 (or the "nearly originial" x^5 - 1 = 0) indicating that 1 was a solution (though this was obvious anyway)...
@Blaqjaqshellaq9 сағат бұрын
Can you figure out the SEVENTH roots of unity? (I dare ya!)