Question

Simplify.
(x + i)(x + i^2)(x i^3)...(x + i^2013)(x + i^2014)(x - i^2014)(x + i^2013)...(x - i^2)(x - i)

Answer 1

i^1=i =i^5 = i^9 ....

i^2=-1 = i^6 = i^10 ....

i^3=-i = i^7 = i^11 ....

i^4=1 = i^8 = i^12

Answer 2

So simplify each of the ` i `s raised to an exponent bigger than 3.

Answer 3

I mean bigger than 1

Answer 4

@SolomonZelman Then what should I do next?

Answer 5

I have a feeling there's a theorem out that says something about simplifying/expanding an \(n\)th degree polynomial if you're given their roots in this kind of factored form...

First thing you can do is to combine some of the factors:
\[\large\color{red}{(x+i)}\color{blue}{(x+i^2)}\cdots\color{blue}{(x-i^2)}\color{red}{(x-i)}\\\large~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\color{red}{(x^2-i^2)}\color{blue}{(x^2-i^4)}\cdots(x^2-i^{4026})(x^2-i^{4028})\]
Now all the exponents on the \(i\)'s are even. You can thus reduce this to
\[\large\underbrace{(x^2-i^2)}_{\text{multiple of 2}}~~\underbrace{(x^2-i^4)}_{\text{multiple of 4}}\cdots\underbrace{(x^2-i^{4026})}_{\text{multiple of 2}}~~\underbrace{(x^2-i^{4028})}_{\text{multiple of 4}}\]
With \(i^2=-1\) and \(i^4=1\), you have
\[\large (x^2+1)(x^2-1)\cdots(x^2+1)(x^2-1)\]
with 2014 factors each of \(x^2+1\) and \(x^2-1\), i.e.
\[\large(x^2+1)^{2014}(x^2-1)^{2014}\]