Solving A Nice Quintic | Problem 477

[@scottleung9587]

Nice!

[@awesome7006]

This is just a 5th roots of unity problem; thus the solutions are e^(0ipi/5), e^(2ipi/5), e^(4ipi/5), e^(6ipi/5), and e^(8ipi/5).

[@Don-Ensley]

problem z⁵ = 1 Use the polar form of 1. 1 = e^(i 2πN), N ∈ ℤ z⁵ = e^(i 2πN) Take to (1/5) power. z = e^(i 2πN/5) The distinct values of N are from 0 to 4, then the angles repeat. N=0, angle 0° z = 1 N=1, angle 72° z = e^(i 2π/5) = (√5-1)/4 + i√(10+2√5)/4 N=2, angle 144° z = e^(i 2πN/5) = -(√5+1)/4 + i √(10-2√5)/4 N=3, angle 216° z = e^(i 2πN/5) = -(√5+1)/4 - i √(10-2√5)/4 N=4, angle 288° z = e^(i 2πN/5) = (√5-1)/4 - i√(10+2√5)/4 answer z ∈ { 1, (√5 - 1) / 4 + i √(10+2√5)/4, -(√5 + 1) / 4 + i √(10-2√5)/4, -(√5 + 1) / 4 - i √(10-2√5)/4, (√5 - 1) / 4 - i √(10+2√5)/4 }

[@mcwulf25]

There's a neat solution for finding cos(2pi/5). Express in terms of cos(pi/5) and then equate cos(4pi/5) to cos(pi - pi/5). All using double angle formulae and a quadratic equation.

[@NadiehFan]

You really should have finished your first method. At 5:53 we have (z² + ¹⁄₂z + k)² = (2k − ³⁄₄)z² + (k − 1)z + (k² − 1) Do you see what I see? People shouldn't believe you when you say that solving this is going to be super time consuming, because it is not. Both the coefficient of the linear term _and_ the constant term of the quadratic in z at the right hand side reduce to zero for k = 1 (making its discriminant equal to zero) and this gives us (z² + ¹⁄₂z + 1)² = ⁵⁄₄z² or (z² + ¹⁄₂z + 1)² = (¹⁄₂√5·z)² which is easy to solve.

[@Etienne-pq3dx]

Why this channel couldn't be (Re) named ? R*e(i*arg(z) ) for example ? 😉 When polynome degree is high, polar form is often more efficient.