2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ – José Carlos Santos Aug 14 '18 at 16:11
  • $\begingroup$ I can't read the question in the link. Please type it using MathJax. $\endgroup$ – saulspatz Aug 14 '18 at 16:15
  • $\begingroup$ Please use informative titles. $\endgroup$ – Did Aug 14 '18 at 16:43
2
$\begingroup$

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 $\,(z-1)^5-1\,$.

$\endgroup$
1
$\begingroup$

$$S=\mathbf(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)$$ $$S(1-\alpha)=\mathbf(1+\alpha^4)(1+\alpha^3)(1+\alpha^2)(1+\alpha)(1-\alpha)$$ $$S(1-\alpha)=(1-\alpha^8)(1+\alpha^3)$$ $$S(1-\alpha)=(1-\alpha^3)(1+\alpha^3)$$ $$S(1-\alpha)=(1-\alpha^6)$$ $$S(1-\alpha)=(1-\alpha)$$ $$S=1$$

$\endgroup$
0
$\begingroup$

After all, you would think there is some trick behind this. There is. Expanding the brackets does not seem to be a solution.

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 $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.