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(D) Let $|z| = r$. Using the triangle inequality for complex numbers,we have $||z| - |1/z|| \leq |z - 1/z| \leq |z| + |1/z|$.Given $|z - 1/z| = 2$,we

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$\sqrt{2} + 1$

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(D) Let $|z| = r$. Using the triangle inequality for complex numbers,we have $||z| - |1/z|| \leq |z - 1/z| \leq |z| + |1/z|$.Given $|z - 1/z| = 2$,we have $|r - 1/r| \leq 2 \leq r + 1/r$.Considering the inequality $|r - 1/r| \leq 2$,we have $-2 \leq r - 1/r \leq 2$.Since $r > 0$,we focus on $r - 1/r \leq 2$,which implies $r^2 - 2r - 1 \leq 0$.Solving the quadratic equation $r^2 - 2r - 1 = 0$ using the quadratic formula,we get $r = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.Since $r = |z| > 0$,the range for $r$ is $0 Therefore,the maximum value of $|z|$ is $1 + \sqrt{2}$.

If $z \neq 0$ is a complex number such that $|z - \frac{1}{z}| = 2$,then the maximum value of $|z|$ is:

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