A bit challenging problem on complex numbers


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}.


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

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