If left|z-frac{2}{z}right|=2,then the greates | vedclass

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(C) Given $\left|z-\frac{2}{z}\right|=2$. Using the triangle inequality property $||z_1|-|z_2|| \leq |z_1-z_2|$,we have: $||z|-\left|\frac{2}{z}\right

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$\sqrt{3}+1$

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(C) Given $\left|z-\frac{2}{z}\right|=2$. Using the triangle inequality property $||z_1|-|z_2|| \leq |z_1-z_2|$,we have: $||z|-\left|\frac{2}{z}\right|| \leq \left|z-\frac{2}{z}\right| = 2$. Let $|z| = r$. Then $|r - \frac{2}{r}| \leq 2$. This implies $-2 \leq r - \frac{2}{r} \leq 2$. Considering the right inequality: $r - \frac{2}{r} \leq 2 \Rightarrow r^2 - 2r - 2 \leq 0$. Solving the quadratic equation $r^2 - 2r - 2 = 0$ using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $r = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$. Since $r = |z| > 0$,we have $r \leq 1 + \sqrt{3}$. Thus,the greatest value of $|z|$ is $\sqrt{3} + 1$.

If $\left|z-\frac{2}{z}\right|=2$,then the greatest value of $|z|$ is

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