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The maximum value of $|x|$, when the complex number $z$ satisfies the condition $\left|z+\frac{2}{z}\right|=2$ is
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Verified Answer
The correct answer is:
$\sqrt{3}+1$
We have the identity,
$$
\begin{aligned}|z|=\left|z+\frac{2}{z}-\frac{2}{z}\right| \leq\left|z+\frac{2}{z}\right|+\left|\frac{2}{z}\right| \\ \Rightarrow \quad|z| \leq 2+\frac{2}{|z|} \end{aligned}
$$
$\Rightarrow \quad|z|^{2} \leq 2|z|+2$
$\Rightarrow \quad|z|^{2}-2|z|+1 \leq 3$
\begin{array}{ll}
\Rightarrow & (|z|-1)^{2} \leq 3 \\
\Rightarrow & -\sqrt{3} \leq|z|-1 \leq \sqrt{3} \\
\hline
\end{array}
$\Rightarrow \quad-\sqrt{3}+1 \leq|z| \leq \sqrt{3}+1$
Hence, the maximum value $=\sqrt{3}+1$
$$
\begin{aligned}|z|=\left|z+\frac{2}{z}-\frac{2}{z}\right| \leq\left|z+\frac{2}{z}\right|+\left|\frac{2}{z}\right| \\ \Rightarrow \quad|z| \leq 2+\frac{2}{|z|} \end{aligned}
$$
$\Rightarrow \quad|z|^{2} \leq 2|z|+2$
$\Rightarrow \quad|z|^{2}-2|z|+1 \leq 3$
\begin{array}{ll}
\Rightarrow & (|z|-1)^{2} \leq 3 \\
\Rightarrow & -\sqrt{3} \leq|z|-1 \leq \sqrt{3} \\
\hline
\end{array}
$\Rightarrow \quad-\sqrt{3}+1 \leq|z| \leq \sqrt{3}+1$
Hence, the maximum value $=\sqrt{3}+1$
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