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:

$$1 - x = \omega \, x, \; \omega^{5} = 1$$

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