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If $\large\frac{1-i}{1+i}$ is a root of the equation $ax^{2}+bx+1=0$ where $a ,b $ are real then $(a , b) $is

\[\begin{array}{1 1}(1)(1 , 1)&(2)(1 , -1)\\(3)(0 , 1)&(4)(1 , 0)\end{array}\]

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The given roots is
$\alpha =\large\frac{1-i}{1+i}$
$\qquad= \large\frac{(1+i)(1-i)}{(1+i)(1-i)}$
$\qquad= \large\frac{1-2i+i^2}{1^2-i^2}$
$\qquad= \large\frac{1-2i}{1+1}$
$\qquad=\large\frac{-21}{2}$$=-i$
$\therefore$ The other root is $ \beta=i$
Sum $\alpha+\beta =-i+i=0$
Product $\alpha \beta = -i \times i =- i^2 =1$
The quadratic equation is
$x^2 $- (sum of the roots)x +product of the roots =0$
$x^2-0x+1=0$
$x^2+1=0$
$a=1,b=0$
$(a,b)=(1,0)$
Hence 4 is the correct answer.
answered May 15, 2014 by meena.p
 

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