Finding solution of equation in complex domain?

[cdummie]

(1+a)ⁿ = (1-a)ⁿ

I tried following:

(1+a)ⁿ = (1-a)ⁿ

[(1+a)/(1-a)]ⁿ=1 but what can i do next?

 

[Mark44]

(1+a)ⁿ = (1-a)ⁿ

I tried following:

(1+a)ⁿ = (1-a)ⁿ

[(1+a)/(1-a)]ⁿ=1 but what can i do next?

The first thing I did was to look at this equation using a couple of small values for n: n = 3 and n = 4, to get some idea of solutions.

Next, I would expand each side using the Binomial Theorem.

 

[mathman]

$(\frac{1+a}{1-a})^n=1$ Has n solutions the n nth roots of unity $e^{\frac {2\pi ki}{n}}=c_k$ for k=0,1,...,n-1.
This leads to $a_k=\frac{c_k-1}{c_k+1}$.

 

[cdummie]

The first thing I did was to look at this equation using a couple of small values for n: n = 3 and n = 4, to get some idea of solutions.

Next, I would expand each side using the Binomial Theorem.

But how can i do that if i don't know the value of a, i mean if i have n=3 then i should first represent both sides in parentheses as cosx + isinx and then find (a+1) and (a-1) to the third degree, but i can't find that trigonometric expression because i don't know the value of a.

 

[mathman]

When n = 3, expanding by the binomial leads to $3a^2+1=0$.