An algebra problem by Madhavan V

$x_{1}$ and $x_{2}$ be roots of $x^{2} +ax + b = 0$ and $\left |x_{1} \right | = \left |x_{2} \right | = 1$

let $x_{1} = x_{2} = p + qi$ ------ (no matter what $p$ equal to $q$ or not )

which $-a = 2p + 2qi$ and $b = \left ( p^{2}-q^{2} \right ) + 2pqi$

$\left |-a \right | =\left |a \right |= 2\sqrt{p^{2} + q^{2}}$ and $\left |b \right | = \sqrt{\left ( p^{2} - q^{2} \right )^{2} + 4p^{2}q^{2}}$

$\left | a \right | = 2$ and $\left |b \right | = \sqrt{p^{4} + q^{4} + 2p^{2}q^{2}} = \sqrt{\left ( p^{2} + q^{2} \right )^{2} } = 1$

leads to

$y^{2} + \left | a \right | y + \left | b \right | = y^{2} + 2 y + 1 = \left (y+1 \right )^{2} =0$

then $\left |y_{1} \right | + \left | y_{2} \right | = \left | -1 \right | + \left | -1 \right | = 2$

Consider a and b as real as all real numbers are complex and apply Vieta's formula

Given $x_1$ , $x_2$ be roots of $x^{2} + ax + b = 0$

$x_1x_2 = b$ (Product of roots) $=> | x_1 |.| x_2 | = | b |$ $=> | b | = | x_1x_2 | = | 1.1 | = 1 ... (1)$

Also, $x_1 + x_2 = -a$ (Sum of roots) $=> | x_1 + x_2 | = | -a | = | a |$ $=> | a |^{2} = | x_1 + x_2 |^{2} = | x_1 |^{2} + | x_2 |^{2} + x_1\bar{x_2} + \bar{x_1}x_2 = 1 + 1 + \frac{x_1}{x_2} + \frac{x_2}{x_1} ... from (1)$ $=> | a |^{2} = 2 + \frac{ x_1^{2} + x_2^{2} }{ x_1x_2 } = 2 + \frac{ (x_1 + x_2)^{2} - 2x_1x_2 }{ x_1x_2 } = 2 + \frac{a^{2} - 2b}{b}$ = \frac{a^{2}}{b}

Now, given $y_1$ , $y_2$ be roots of $y^{2} + |a|y + |b| = 0$

$y_1y_2 = |b| = 1 ... from (1)$

$| y_1 + y_2 | = | - |a| | = | a |$ $=> | y_1 |^{2} + | y_2 |^{2} = | a |^{2} - y_1\bar{y_2} - \bar{y_1}y_2 = \frac{a^{2}}{b} - \frac{y_1}{\bar{y_1}} - \frac{y_2}{\bar{y_2}} = 2$