The maximum value of |z| where z satisfies th | vedclass

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(B) Given $\left| z + \frac{2}{z} \right| = 2$. Using the triangle inequality,we know that $\left| z + \frac{2}{z} \right| \ge ||z| - |\frac{2}{z}||$.

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$\sqrt{3} + 1$

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(B) Given $\left| z + \frac{2}{z} \right| = 2$. Using the triangle inequality,we know that $\left| z + \frac{2}{z} \right| \ge ||z| - |\frac{2}{z}||$. Also,by the reverse triangle inequality,$|z + \frac{2}{z}| \le |z| + |\frac{2}{z}|$. From the given condition,we have $|z| - \frac{2}{|z|} \le |z + \frac{2}{z}| = 2$. Let $|z| = r$. Then $r - \frac{2}{r} \le 2$,which implies $r^2 - 2r - 2 \le 0$. Solving the quadratic equation $r^2 - 2r - 2 = 0$,we get $r = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$. Since $r = |z| > 0$,we have $r \le 1 + \sqrt{3}$. Thus,the maximum value of $|z|$ is $1 + \sqrt{3}$.

The maximum value of $|z|$ where $z$ satisfies the condition $\left| z + \frac{2}{z} \right| = 2$ is

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