Find the fifth roots of 1?

answers:
Φ² = Φ+1 say: Find the fifth roots of 1
z⁵ = 1
Solution 1  Geometry: Clearly (1)⁵ = 1, so z = 1 is a solution. In the complex plane, this and the remaining four solutions form the vertices of a regular pentagram (from "⁵") circumscribed by the unit circle that is centred at the origin (0+0i). A little trig later you find the five solutions are: 1 (which is 1+0i or cos(π)+sin(π)i), cos(⁷∕₅π)+sin(⁷∕₅π)i, cos(⁹∕₅π)+sin(⁹∕₅π)i, cos(⅕π)+sin(⅕π)i, cos(⅗π)+sin(⅗π)i. Evaluate or approximate as required.
Solution 2  Algebra: z⁵ = 1, so z⁵ + 1 = 0. This factors to (z + 1)(z⁴  z³ + z²  z + 1) = 0
to (z + 1)(z²  ((1+√5)/2)z + 1)(z² + ((1+√5)/2)z + 1) = 0
to (z + 1)(z  (√5+1)/4 + i√(2(5√5))/4)(z  (√5+1)/4  i√(2(5√5))/4)(z + (√51)/4  i√(2(5+√5))/4)(z + (√51)/4 + i√(2(5+√5))/4) = 0
So the five solutions are: 1+0i, (√5+1)/4  i√(2(5√5))/4, (√5+1)/4 + i√(2(5√5))/4, (1√5)/4 + i√(2(5+√5))/4, and (1√5)/4  i√(2(5+√5))/4

sepia say: The fifth root of 1:
0.80901699437... + 0.587785252292473... i

rotchm say: 1 = cos(π) + isin(π) = e^(i(π + 2nπ)). Why?
Thus the 5 roots of 1 are (e^(i(π + 2nπ)))^(1/5) = e^(i(π + 2nπ)/5).
So, what are the five angles (π + 2nπ)/5 ?
Done!

Smith say: It's 1, because (1) * (1) * (1) * (1) * (1) = 1

Morningfox say: Those would be, in polar form:
1 @ 180 deg
1 @ 36 deg
1 @ 108 deg
1 @ 252 deg
1 @ 354 deg
I'm sure you can convert to Cartesian (x, y) system if you need to do that.

Captain Matticus, LandPiratesInc say: 1 =>
1 + 0i =>
R * cos(t) + R * sin(t) * i
R * cos(t) = 1
R * sin(t) = 0
R^2 * cos(t)^2 + R^2 * sin(t)^2 = (1)^2 + 0^2
R^2 * (cos(t)^2 + sin(t)^2) = 1
R^2 * 1 = 1
R^2 = 1
R = 1 ,1
Let's just assume that R = 1 for the sake of simplicity
cos(t) = 1
t = pi + 2pi * k, where k is an integer
sin(t) = 0
t = pi * k, where k is an integer
The most restrictive solution set is t = pi + 2pi * k
1 =>
1 + 0i =>
cos(pi + 2pi * k) + i * sin(pi + 2pi * k) =>
cos(pi * (1 + 2k)) + i * sin(pi * (1 + 2k))
DeMoivre's Theorem tells us that (cos(t) + i * sin(t))^n = cos(n * t) + i * sin(n * t)
(1)^(1/5) =>
(cos(pi * (1 + 2k)) + i * sin(pi * (1 + 2k)))^(1/5) =>
cos((pi/5) * (1 + 2k)) + i * sin((pi/5) * (1 + 2k))
Let's restrict t between 0 and 2pi. This will give us a complete set of roots
0 < (pi/5) * (1 + 2k) < 2pi
0 < 1 + 2k < 10
1 < 2k < 9
1/2 < k < 9/2
k is an integer
k = 0 , 1 , 2 , 3 , 4
(pi/5) * (1 + 2k)
(pi/5) * (1 + 0) , (pi/5) * (1 + 2) , (pi/5) * (1 + 4) , (pi/5) * (1 + 6) , (pi/5) * (1 + 8) =>
pi/5 , 3pi/5 , 5pi/5 , 7pi/5 , 9pi/5 =>
pi/5 , 3pi/5 , pi , 7pi/5 , 9pi/5
cos(pi/5) + i * sin(pi/5)
cos(3pi/5) + i * sin(3pi/5)
cos(pi) + i * sin(pi)
cos(7pi/5) + i * sin(7pi/5)
cos(9pi/5) + i * sin(9pi/5)
We can use a bunch of trig identities to find the exact values for each angle, or we could use a reference table. At this point, we're fine. All of the roots have been found.
https://en.wikipedia.org/wiki/Trigonomet...

Aster Rhoids say: It's the same as 1^(1/5)
