If z is a complex number such that left|z-fra | vedclass

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

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$\sqrt{5}+1$

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(D) Given,$\left|z-\frac{4}{z}\right|=2$.Using the triangle inequality $|a+b| \leq |a|+|b|$,we have:$|z| = \left|z-\frac{4}{z} + \frac{4}{z}\right| \leq \left|z-\frac{4}{z}\right| + \left|\frac{4}{z}\right|$.Substituting the given value: $|z| \leq 2 + \frac{4}{|z|}$.Multiplying by $|z|$ (since $|z| > 0$): $|z|^2 \leq 2|z| + 4$.$|z|^2 - 2|z| - 4 \leq 0$.Completing the square: $(|z|-1)^2 - 1 - 4 \leq 0 \Rightarrow (|z|-1)^2 \leq 5$.Taking the square root: $|z|-1 \leq \sqrt{5}$.Therefore,$|z| \leq \sqrt{5}+1$.The greatest value of $|z|$ is $\sqrt{5}+1$.

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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