# Homework Help: Find roots of complex equation (1-x)^5 = x^5

1. Sep 6, 2011

### DryRun

1. The problem statement, all variables and given/known data
Find roots of complex equation (1-x)^5 = x^5

2. Relevant equations
Probably Euler and/or De Moivre.

3. 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).

2. Sep 6, 2011

### micromass

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.

$$1 - x = \omega \, x, \; \omega^{5} = 1$$
You get a linear equation in $x$ parametrized by the number $\omega$. Solve it for $x$. What are the possible values for $\omega$?