## A bit challenging problem on complex numbers

Problem:

If $\omega$ and $\omega^{2}$ satisfy the equation

$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x} = \frac{2}{x}$

then find the value of $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1}$.

Solution:

We can write the given equation as

$\sum {x(b+x)(c+x)(d+x)}=2(a+x)(b+x)(c+x)(d+x)$

$\Longrightarrow \sum {x(x^{3}+(b+c+d)x^{2}+(bc+cd+bd)x+bcd)}$

$= 2{(x^{4}+(a+b+c+d))x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+abd+acd+bcd)x+abcd)}$

$\Longrightarrow 2x^{4}+(a+b+c+d)x^{3}+0x^{2}-(abc+abd+acd+bcd)x-2abcd=0$

This is a fourth degree equation whose two roots are $\omega, \omega^{2}$. Let $\alpha, \beta$ be the other two roots. Then,

$(\alpha + \beta)(\omega + \omega^{2}) + \alpha \beta + \omega . \omega^{2} = 0$ (sum of the products taken two at a time)

$\Longrightarrow (\alpha + \beta)(-1)+\alpha \beta +1 =0$

$\Longrightarrow (1 - \alpha)(1 - \beta) = 0$

$\alpha = 1$ or $\beta = 1$

Hence, we get $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1}=1$.

Note that in deriving the fourth degree equation we used a basic technique of expansion or multiplication of several binomials.

More later,

Nalin Pithwa