## Complex ain’t so complex ! Learning to think!

Problem:

If 1, $\omega, \omega^{2}, \omega^{3}, \ldots, \omega^{n}$ are the nth roots of unity, then find the value of $(2-\omega)(2-\omega^{2})(2-\omega^{3})\ldots (2-\omega^{n-1})$.

Solution:

Learning to think:

Compare it with what we know from our higher algebra — suppose we have to multiply out:

$(x+a)(x+b)(x+c)(x+d)$. We know it is equal to the following:

$x^{4}+(a+b+c+d)x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+acd+bcd+abd)x+abcd$

If we examine the way in which the partial products are formed, we see that

(1) the term $x^{4}$ is formed by taking the letter x out of each of the factors.

(2) the terms involving $x^{3}$ are formed by taking the letter x out of any three factors, in every possible way, and one of the letters a, b, c, d out of the remaining factor

(3) the terms involving $x^{2}$ are formed by taking the letter x out of any two factors, in every possible way, and two of the letters a, b, c, d out of the remaining factors

(4) the terms involving x are formed by taking the letter x out of any one factor, and three out of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters a, b, c, d.

Further hint:

relate the above to sum of binomial coefficients.

and, you are almost done.

More later,

Nalin Pithwa

### 2 Comments

1. sachit.shanbhag1+NP@gmail.com
Posted February 15, 2016 at 5:27 pm | Permalink | Reply

Roots of the polynomial (2-x)^n=1 will be 2-w , 2-w^2 …
Hence product of the numbers will be 2^n-1 or 1-2^n depending on the value of n(It will decide the coefficient of x^n.

• Posted February 16, 2016 at 11:11 am | Permalink | Reply

I agree that the root of $(2-\omega)^{n}=1$ is $2-\omega$. But, $(2-\omega^{2})$ does not seem obviously the root of $(2-\omega)^{2}=1$. Any way, what is the final answer as per your reasoning?

This site uses Akismet to reduce spam. Learn how your comment data is processed.