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(B) Given $\left| \omega + \frac{1}{\omega} \right| = 2$. Using the triangle inequality $\left| z_1 + z_2 \right| \le \left| z_1 \right| + \left| z_2

Vedclass · Accepted Answer

$1 + \sqrt{2}$

Vedclass · Answer

(B) Given $\left| \omega + \frac{1}{\omega} \right| = 2$. Using the triangle inequality $\left| z_1 + z_2 \right| \le \left| z_1 \right| + \left| z_2 \right|$,we have: $\left| \omega \right| = \left| \left( \omega + \frac{1}{\omega} \right) - \frac{1}{\omega} \right| \le \left| \omega + \frac{1}{\omega} \right| + \left| \frac{1}{\omega} \right|$. Substituting the given value: $\left| \omega \right| \le 2 + \frac{1}{\left| \omega \right|}$. Let $r = \left| \omega \right|$. Then $r \le 2 + \frac{1}{r}$,which implies $r^2 - 2r - 1 \le 0$. Solving the quadratic equation $r^2 - 2r - 1 = 0$ using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $r = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$. Since $r = \left| \omega \right| > 0$,we take $r \le 1 + \sqrt{2}$. Thus,the maximum distance of $\omega$ from the origin is $1 + \sqrt{2}$.

If $\omega$ is a complex number satisfying $\left| \omega + \frac{1}{\omega} \right| = 2$,then the maximum distance of $\omega$ from the origin is:

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