Tag Archives: nth roots of unity

Complex attitude!

Let us a discuss yet one more complex number based IITJEE mains problem.

Problem:

Let z_{1} and z_{2} be nth roots of unity, which subtend a right angle at the origin. Then, integer n must be of the form __________________. (fill in the blank).

Solution:

nth roots of unity are given by

\cos{\frac{2m\pi}{n}}+ i \sin{\frac{2m\pi}{n}}= e^{\frac{2m\pi}{n}}

where m=0,1,2, \ldots, n-1.

Let z_{1}=\cos{\frac{2m_{1}\pi}{n}}+ i\sin{\frac{2m_{1}\pi}{n}}=e^{2m_{1}\pi i/n}

Let z_{2}=e^{2m_{2}\pi i/n} where 0 \leq m_{1}, m_{2}< n, m_{1} \neq m_{2}

As the join of z_{1} and z_{2} subtends a right angle at the origin, we deduce that \frac{z_{1}}{z_{2}} is purely imaginary.

\Longrightarrow \frac{e^{2m_{1}\pi i/n}}{e^{2m_{2}\pi i/n}}=ik, for some real k

\Longrightarrow e^{2(m_{1}-m_{2})\pi i/n}=ik

\Longrightarrow n=4(m_{1}-m_{2}). Thus, n must be of the form 4k.

More later,

Nalin Pithwa

 

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