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

Answer. (a).

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: