Find roots of complex equation (1-x)^5 = x^5

[DryRun]

Homework Statement

Find roots of complex equation (1-x)^5 = x^5

Homework Equations

Probably Euler and/or De Moivre.

The Attempt at a Solution

I know i need to have z^5 on one side and all the rest on the other side. But i need 5 roots, and I'm only getting one.

(1-x)^5 = x^5
Use fifth root on both sides.
(1-x) = x
2x= 1
x = 1/2 (only one solution, which i know is wrong, but no idea how to proceed).

 

[micromass]

Homework Statement

Find roots of complex equation (1-x)^5 = x^5

Homework Equations

Probably Euler and/or De Moivre.

The Attempt at a Solution

I know i need to have z^5 on one side and all the rest on the other side. But i need 5 roots, and I'm only getting one.

(1-x)^5 = x^5
Use fifth root on both sides.
(1-x) = x
2x= 1
x = 1/2 (only one solution, which i know is wrong, but no idea how to proceed).

You have the right idea here. But (like you thought), it isn't true that a^5=b^5 implies a=b. However, we can fix this:

If a^5=b^5 (and they are nonzero), then \left(\frac{a}{b}\right)^5=1. So a/b are fifth roots of unity. There are 5 fifth roots of unity. So let \alpha_i be one of the fifth roots of unity, then we have

\frac{a}{b}=\alpha_i

And thus

a=\alpha_i b

So, if a^5=b^5, then a=\alpha_i b where we take \alpha_i the fifth roots of unity. This gives us 5 equations.

Try that on your equation.

 

[Dickfore]

Take the complex root, to get:

<br /> 1 - x = \omega \, x, \; \omega^{5} = 1<br />

You get a linear equation in x parametrized by the number \omega. Solve it for x. What are the possible values for \omega?