# 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

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

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 $\alpha \beta = -i \times i =- i^2 =1$