# Compute $(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)$ where $\alpha$ is the complex 5th root of unity with the smallest positive principal argument

I just started on the topic Complex Numbers and there is a question that I am stuck on.

The question is:

If $\alpha$ is a complex 5th root of unity with the smallest positive principal argument, determine the value of $\mathbf(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)$

From what I understand, I'm supposed to start with a^5=1 and that the smallest positive argument should be 2π/5 To be exact, I got the roots $\mathbf{e^{\frac{2ki\pi}{5}}}$ in which k is from 0 to 4. After that, I have no idea how to proceed.

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Aug 14 '18 at 16:11
• I can't read the question in the link. Please type it using MathJax. – saulspatz Aug 14 '18 at 16:15
• Please use informative titles. – Did Aug 14 '18 at 16:43

Hint: the five $5^{th}$ roots of unity are $\,1,\alpha,\alpha^2,\alpha^3,\alpha^4\,$. Also, if $\,\beta\,$ is a root of $\,z^5-1\,$ then $\,1+\beta\,$ is a root of $\,(z-1)^5-1\,$, and therefore $\,(1+1)(1+\alpha)(1+\alpha^2)(1+\alpha^3)(1+\alpha^4)\,$ is the product of the five roots of $\,(z-1)^5-1\,$.
$$S=\mathbf(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)$$ $$S(1-\alpha)=\mathbf(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)(1-\alpha)$$ $$S(1-\alpha)=(1-\alpha^8)(1+\alpha^3)$$ $$S(1-\alpha)=(1-\alpha^3)(1+\alpha^3)$$ $$S(1-\alpha)=(1-\alpha^6)$$ $$S(1-\alpha)=(1-\alpha)$$ $$S=1$$
The idea, is that the product of many terms, is (times $(-1)^{\mbox{deg}}$)the constant term of the smallest degree monic polynomial of which they are roots. This is an example of Vieta's formula.
Here, note that if $x$ is any one of $1 + \alpha^k$, $1 \leq k \leq 5$ then $(x - 1)^5 = 1$, or $(x-1)^5 - 1 = 0$.
That is, there are five roots of $(x-1)^5 - 1 = 0$, and these are all the roots by the fundamental theorem of algebra. The product of the roots is $(1 + \alpha)(1 + \alpha^2)(1+\alpha^3)(1+\alpha^4)(1+1)$, and this is equal to (by Vieta's formula) the negative of the constant term of the above polynomial, which is $(-1)(-1^5 - 1) = 2$ by the binomial theorem. Dividing by $(1+1) = 2$ gives the desired product as $1$.