Finding a complex number from an equation

Temp0
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Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.
 
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Temp0 said:

Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.

Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?
 
We used to solve these in a geometrically way noting that the roots will be on a circle equally spaced at every 60 degrees about 0+0i and with a radius of 2. So given one root you can find the others.
 
Last edited:
Dick said:
Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?

Well, r^6 = 64 so r = 2, and then because a / r = - 32, and b / r = 0, I just assumed that the angle I start at is pi, so my angle equation becomes 6θ = ∏ + 2∏k, and oh wait I pretty much just found my error as soon as I typed it out LOL, thanks alot.
 
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