Key Concepts and Formulas

• Reverse Triangle Inequality: For complex numbers $z_1$ and $z_2$, $||z_1| - |z_2|| \le |z_1 - z_2|$.
• Modulus of Reciprocal: For a non-zero complex number $z$, $|\frac{1}{z}| = \frac{1}{|z|}$.
• Absolute Value Inequality: $|A| \le B$ is equivalent to $-B \le A \le B$.

Step-by-Step Solution

1. Applying the Reverse Triangle Inequality: We apply the reverse triangle inequality to the given expression $|z - \frac{1}{z}|$. This allows us to relate the magnitude of the difference to the individual magnitudes. $||z| - \left|\frac{1}{z}\right|| \le \left|z - \frac{1}{z}\right|$

2. Simplifying the Modulus of $\frac{1}{z}$: We use the property that the modulus of the reciprocal is the reciprocal of the modulus. This simplifies the expression inside the absolute value. $\left||z| - \frac{1}{|z|}\right| \le \left|z - \frac{1}{z}\right|$

3. Using the Given Condition: We substitute the given value $\left|z - \frac{1}{z}\right| = 2$ into the inequality. This sets up the inequality we need to solve for $|z|$. $\left||z| - \frac{1}{|z|}\right| \le 2$

4. Substituting $x = |z|$: Let $x = |z|$. Since $z \neq 0$, we have $x > 0$. This substitution simplifies the notation and makes the algebraic manipulations clearer. $\left|x - \frac{1}{x}\right| \le 2$

5. Converting to a Compound Inequality: We rewrite the absolute value inequality as a compound inequality. This allows us to work with linear inequalities. $-2 \le x - \frac{1}{x} \le 2$

6. Solving the First Inequality: $x - \frac{1}{x} \le 2$ Multiplying by $x$ (since $x > 0$) gives $x^2 - 1 \le 2x$, which rearranges to $x^2 - 2x - 1 \le 0$. We solve this quadratic inequality. $x^2 - 2x - 1 \le 0$ The roots of $x^2 - 2x - 1 = 0$ are $x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$. Since the parabola opens upwards, the inequality holds between the roots. Thus, $1 - \sqrt{2} \le x \le 1 + \sqrt{2}$. Since $x > 0$, we have $0 < x \le 1 + \sqrt{2}$. $0 < x \le 1 + \sqrt{2} \quad \text{(Equation 1)}$

7. Solving the Second Inequality: $-2 \le x - \frac{1}{x}$ Rearranging, we get $x - \frac{1}{x} + 2 \ge 0$. Multiplying by $x$ (since $x > 0$) gives $x^2 + 2x - 1 \ge 0$. We solve this quadratic inequality. $x^2 + 2x - 1 \ge 0$ The roots of $x^2 + 2x - 1 = 0$ are $x = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$. Since the parabola opens upwards, the inequality holds outside the roots. Thus, $x \le -1 - \sqrt{2}$ or $x \ge -1 + \sqrt{2}$. Since $x > 0$, we have $x \ge -1 + \sqrt{2}$. $x \ge \sqrt{2} - 1 \quad \text{(Equation 2)}$

8. Combining the Results: We need to satisfy both inequalities. Equation 1 gives $0 < x \le 1 + \sqrt{2}$, and Equation 2 gives $x \ge \sqrt{2} - 1$. Combining these, we have $\sqrt{2} - 1 \le x \le 1 + \sqrt{2}$. Therefore, the maximum value of $x = |z|$ is $1 + \sqrt{2}$. $\sqrt{2} - 1 \le |z| \le \sqrt{2} + 1$

Common Mistakes & Tips

• Forgetting $|z|>0$: Always remember that $|z|$ is a magnitude and therefore must be positive.
• Sign Errors: Be careful when multiplying inequalities by variables; ensure you know the sign.
• Quadratic Inequality Solution: Remember to consider the shape of the parabola (upward or downward opening) when solving quadratic inequalities.

Summary

By applying the Reverse Triangle Inequality and simplifying the resulting expression, we were able to find the range for $|z|$. The maximum value of $|z|$ is found to be $1 + \sqrt{2}$.

The final answer is $\boxed{\sqrt{2}+1}$, which corresponds to option (D).