## 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