The maximum value of |z|, when the complex nu | vedclass

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

Vedclass · Accepted Answer

$\sqrt{3}+1$

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(C) Given the condition $\left|z+\frac{2}{z}\right|=2$. Using the triangle inequality, $|z| = \left|z+\frac{2}{z}-\frac{2}{z}\right| \leq \left|z+\frac{2}{z}\right| + \left|-\frac{2}{z}\right|$. Substituting the given value, $|z| \leq 2 + \frac{2}{|z|}$. Multiplying by $|z|$ (since $|z| > 0$), we get $|z|^2 \leq 2|z| + 2$. Rearranging the terms, $|z|^2 - 2|z| \leq 2$. Completing the square, $|z|^2 - 2|z| + 1 \leq 2 + 1$, which gives $(|z|-1)^2 \leq 3$. Taking the square root, $-\sqrt{3} \leq |z|-1 \leq \sqrt{3}$. Adding $1$ to all sides, $1-\sqrt{3} \leq |z| \leq 1+\sqrt{3}$. Since $|z| \geq 0$, the maximum value of $|z|$ is $1+\sqrt{3}$.

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

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