If left|z+frac{2}{z}right|=2,then the maximum | vedclass

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Question

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

Vedclass · Accepted Answer

$1+\sqrt{3}$

Vedclass · Answer

(A) Given,$\left|z+\frac{2}{z}\right|=2$. Using the triangle inequality,$|z| = |(z+\frac{2}{z}) - \frac{2}{z}| \leq |z+\frac{2}{z}| + |\frac{2}{z}|$. Substituting the given value,$|z| \leq 2 + \frac{2}{|z|}$. Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 \leq 2|z| + 2$,which implies $|z|^2 - 2|z| - 2 \leq 0$. Solving the quadratic inequality $x^2 - 2x - 2 \leq 0$ for $x = |z|$,the roots are $x = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$. Since $|z| \geq 0$,we have $0 \leq |z| \leq 1+\sqrt{3}$. Thus,the maximum value of $|z|$ is $1+\sqrt{3}$.

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

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