Find the fifth roots of -1?

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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 + (√5-1)/4 - i√(2(5+√5))/4)(z + (√5-1)/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
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sepia say: The fifth root of -1:
0.80901699437... + 0.587785252292473... i
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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!
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Smith say: It's -1, because (-1) * (-1) * (-1) * (-1) * (-1) = -1
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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.
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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...
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Aster Rhoids say: It's the same as -1^(1/5)
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