What are the six (complex) solutions to y=64

clearly 2 and -2 are 2 of the six

I am assuming you want the six complex solutions to:
y^6 = 64.

Note that the solutions for y are the six sixth roots of 64. To find the sixth roots of 64, re-write 64 in polar form to get:
64 = 64(cos 0 + i sin 0).

By DeMovire's Theorem:
y = 64^(1/6)
= [64(cos 0 + i sin 0)]^(1/6)
= 2{cos[(0 + 2πk)/6] + i sin[(0 + 2πk)/6]}, where k = 0, 1, 2, 3, 4, and 5
= 2[cos(πk/3) + i sin(πk/3)].

Therefore, by letting k = 0, 1, 2, 3, 4, and 5, the required complex roots are:
(a) k = 0 ==> y = 2(cos 0 + i sin 0) = 2, which is an obvious one
(b) k = 1 ==> y = 2(cos π/3 + i sin π/3) = 1 + i√3
(c) k = 2 ==> y = 2(cos 2π/3 + i sin 2π/3) = -1 + i√3
(d) k = 3 ==> y = 2(cos π + i sin π) = -2, the other obvious one
(e) k = 4 ==> y = 2(cos 4π/3 + i sin 4π/3) = -1 - i√3
(f) k = 5 ==> y = 2(cos 5π/3 + i sin 5π/3) = 1 - i√3.

I hope this helps!

The one and only solution to y = 64 is y = 64.

Did you mean y^6 = 64?

Solutions have magnitude 2 and have angles 0, π/3, 2π/3, π, 4π/3 and 5π/3, so the six roots are

y = 2 cis(nπ/3) where n runs from 0 to 5

In rectangular form, they are 2, –2, 1 + i√(3), 1 – i√(3), –1 + i√(3), –1 – i√(3)

I think the equation is meant to be y^6 = 64.
Let cis(theta) denote cos(theta) + i sin(theta).
We have 64 = 64cis(2npi).

To take the 6th roots, find the 6th root of the length and divide the angles by 6.
The 6th roots of 64 are 64^(1/6) * cis(2n pi/6) = 2cis(n pi/3), for n = 0, 1, 2, 3, 4, 5.
So the 6th roots of 64 are

2cis(0pi/3) = 2(cos 0 + i sin 0) = 2,
2cis(1pi/3) = 2(cos (pi/3) + i sin (pi/3)) = 1 + i sqrt(3),
2cis(2pi/3) = 2(cos (2pi/3) + i sin (2pi/3)) = -1 + i sqrt(3),
2cis(3pi/3) = 2(cos pi + i sin pi) = -2,
2cis(4pi/3) = 2(cos (4pi/3) + i sin (4pi/3)) = -1 - i sqrt(3), and
2cis(5pi/3) = 2(cos (5pi/3) + i sin (5pi/3)) = 1 - i sqrt(3).

Lord bless you today!

What is your equation?