Problem:
If 1, are the nth roots of unity, then find the value of
.
Solution:
Learning to think:
Compare it with what we know from our higher algebra — suppose we have to multiply out:
. We know it is equal to the following:
If we examine the way in which the partial products are formed, we see that
(1) the term is formed by taking the letter x out of each of the factors.
(2) the terms involving 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 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
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.
I agree that the root of
is
. But,
does not seem obviously the root of
. Any way, what is the final answer as per your reasoning?