## Complex Numbers — a typical IITJEE Main problem

Example 1.

If $\omega$ is the imaginary cube root of unity, then value of the expression

$1(2-\omega)(2-\omega^{2}) + 2(3-\omega)(3-\omega^{2})+\ldots+(n-1)(n-\omega)(n-\omega^{2})$ is

(a) $\frac{1}{4}{n^{2}}(n+1)^{2}-n$

(b) $\frac{1}{4}{n^{2}}(n-1)^{2}+n$

(c) $\frac{1}{4}{n^{2}}(n+1)-n$

(d) $\frac{1}{4}n(n+1)^{2}-n$

Solution:

rth term of the given expression is

$r(r+1-\omega)(r+1-\omega^{2})=(r+1-1)(r+1-\omega)(r+1-\omega^{2})=(r+1)^{3}-1$

because $x^{3}-1=(x-1)(x-\omega)(x-\omega^{2})$.

Thus, the value of the expression is given by

$\sum_{r=1}^{n-1}r(r+1-\omega)(r+1-\omega^{2})=\sum_{r=1}^{n-1}[(r+1)^{3}-1]$

$=2^{3}+3^{3}+ \ldots + n^{3}-(n-1)$

$=1^{3}+2^{3}+\ldots + n^{3}-n$

$=\frac{1}{4}n^{2}{(n+1)}^{2}-n$

More later,

Nalin Pithwa

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