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Let $$x_1, x_2$$ are the roots of the 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 $$y^2 + \mid a\mid y + \mid b \mid = 0$$. If $$\mid x_1\mid = \mid x_2\mid = 1 $$ then prove that $$\mid y_1\mid = \mid y_2\mid =1$$


Solution

Suppose $$x^2 + ax + b = 0 $$ has root $$x_1\space x_2$$. Then,
$$\therefore x_1 + x_2 = -a$$
and
$$x_1x_2 = b$$
From (2),

$$|x_1| |x_2| = |b|$$
$$\Rightarrow |b| = 1$$
Also,

$$|-a| = |x_1 + x_2|$$

$$\Rightarrow |a| \leq |x_1| + |x_2|$$

or

$$|a| \leq 2$$

Now suppose $$y^2 + |a|y + |b| = 0$$ has roots $$y_1\space and\space y_2$$ Then,

$$y_1=y_2 =\dfrac{ -|a|\pm\sqrt{|a|^2 - 4|b|}}{2}$$

              $$= \dfrac{-|a|\pm(\sqrt{4 - |a|^2})i}{2}$$

$$\Rightarrow |y_1|= |y_2| = \dfrac{\sqrt{|a|^2 + 4 - |a|^2}}{2} = 1$$

Hence, $$|y_1| = |y_2| = 1$$

Mathematics

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