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Question

Let $$x_1, x_2,$$ are the roots of quadratic equation $$x^2 + ax + b = 0$$, Where $$ a, b$$ are complex numbers and $$y_1, y_2$$ are the roots of the quadratic equation $$y^2 + \left|a\right| y + \left|b\right| =0$$. If $$\left|x_1\right| = \left|x_2\right| = 1$$, then 


A
|y1|=|y2|=1.
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B
|y1|=|y2|1.
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C
|y1|1,|y2|=1.
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D
|y1|=1,|y2|1.
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Solution

The correct option is A $$\left|y_1\right| = \left|y_2\right|= 1$$.
Now 
$$x_{1}.x_{2}=b$$
Hence
$$|x_{1}.x_{2}|$$
$$=|x_{1}|.|x_{2}|$$
$$=1.1$$
$$=1$$
$$=|b|$$ ...(i)
And 
$$-a=x_{1}+x_{2}$$
$$|a|=|x_{1}+x_{2}|$$
Now applying triangle inequality, we get 
$$|x_{1}+x_{2}|\leq |x_{1}|+|x_{2}|$$
Hence
$$|a|\leq |x_{1}|+|x_{2}|$$
Hence
$$|a|\leq 2$$
Considering 
$$|a|=2$$ we get 
$$y^{2}+2y+1=0$$
$$(y+1)^{2}=0$$
$$y=-1,-1$$
Hence
$$|y_{1}|=|y_{2}|=1$$.

Mathematics

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