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(D) Given,$\left|z-\frac{4}{z}\right|=2$.Using the triangle inequality,$|z| = \left|z-\frac{4}{z}+\frac{4}{z}\right| \leq \left|z-\frac{4}{z}\right| +

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$1+\sqrt{5}$

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(D) Given,$\left|z-\frac{4}{z}\right|=2$.Using the triangle inequality,$|z| = \left|z-\frac{4}{z}+\frac{4}{z}\right| \leq \left|z-\frac{4}{z}\right| + \left|\frac{4}{z}\right|$.Substituting the given value,$|z| \leq 2 + \frac{4}{|z|}$.Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 - 2|z| - 4 \leq 0$.Solving the quadratic inequality $|z|^2 - 2|z| - 4 = 0$ using the quadratic formula,$|z| = \frac{2 \pm \sqrt{4 - 4(1)(-4)}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}$.Since $|z| > 0$,we have $0 Therefore,the greatest value of $|z|$ is $1 + \sqrt{5}$.

If $z$ is a complex number such that $\left|z-\frac{4}{z}\right|=2$,then the greatest value of $|z|$ is

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