1. Key Concepts and Formulas

• Triangle Inequality: For complex numbers $z_1$ and $z_2$, $|z_1 + z_2| \le |z_1| + |z_2|$.
• Modulus of a Quotient: For complex numbers $z_1$ and $z_2$, $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

2. Step-by-Step Solution

Step 1: Expressing $z$ and Applying the Triangle Inequality

• What: We want to relate $|z|$ to the given condition $|z - \frac{4}{z}| = 2$. To do this, we express $z$ as a sum involving the term $z - \frac{4}{z}$, and then apply the triangle inequality.
• Why: This allows us to use the given information and establish an inequality involving only $|z|$. We can write $z$ as: $z = \left( z - \frac{4}{z} \right) + \frac{4}{z}$ Taking the modulus of both sides: $|z| = \left| \left( z - \frac{4}{z} \right) + \frac{4}{z} \right|$ Applying the triangle inequality, $|z_1 + z_2| \le |z_1| + |z_2|$, with $z_1 = \left( z - \frac{4}{z} \right)$ and $z_2 = \frac{4}{z}$: $|z| \le \left| z - \frac{4}{z} \right| + \left| \frac{4}{z} \right|$ Using the modulus property $\left| \frac{k}{w} \right| = \frac{|k|}{|w|}$, we can simplify the second term: $\left| \frac{4}{z} \right| = \frac{|4|}{|z|} = \frac{4}{|z|}$ Substituting this back into our inequality: $|z| \le \left| z - \frac{4}{z} \right| + \frac{4}{|z|}$

Step 2: Substituting the Given Condition

• What: Substitute the given condition $\left| z - \frac{4}{z} \right| = 2$ into the inequality derived in the previous step.
• Why: This directly incorporates the given information, leading to an inequality involving only $|z|$. $|z| \le 2 + \frac{4}{|z|}$

Step 3: Rearranging into a Quadratic Inequality

• What: We will rearrange the inequality into a standard quadratic form to solve for $|z|$. Let $x = |z|$.
• Why: Solving quadratic inequalities is a standard mathematical technique. By transforming our inequality into this form, we can find the range of possible values for $|z|$. Since $|z|$ represents a modulus, it must be a non-negative real number, i.e., $x \ge 0$. The inequality becomes: $x \le 2 + \frac{4}{x}$ Since $x = |z|$ must be positive (if $|z|=0$, then $\frac{4}{z}$ is undefined), we can multiply both sides by $x$ without changing the direction of the inequality: $x^2 \le 2x + 4$ Rearranging the terms to form a quadratic inequality: $x^2 - 2x - 4 \le 0$

Step 4: Solving the Quadratic Inequality

• What: Find the roots of the quadratic equation $x^2 - 2x - 4 = 0$ using the quadratic formula, and then determine the interval for $x$ that satisfies the inequality.
• Why: The roots of a quadratic equation are the values where the quadratic expression equals zero. For a quadratic inequality of the form $ax^2 + bx + c \le 0$ (with $a > 0$), the values of $x$ that satisfy the inequality lie between these roots. Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, for $x^2 - 2x - 4 = 0$ (where $a=1, b=-2, c=-4$): $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 16}}{2}$ $x = \frac{2 \pm \sqrt{20}}{2}$ $x = \frac{2 \pm 2\sqrt{5}}{2}$ $x = 1 \pm \sqrt{5}$ So, the two roots are $x_1 = 1 - \sqrt{5}$ and $x_2 = 1 + \sqrt{5}$. Since the coefficient of $x^2$ is positive (1 > 0), the parabola opens upwards. For the inequality $x^2 - 2x - 4 \le 0$ to hold, $x$ must lie between or be equal to the roots: $1 - \sqrt{5} \le x \le 1 + \sqrt{5}$

Step 5: Determining the Maximum Value of $\,\left| z \right|$

• What: Substitute back $|z|$ for $x$ and consider the fundamental property that a modulus must be non-negative.
• Why: The modulus of a complex number, by definition, must be greater than or equal to zero. The lower bound obtained from the quadratic inequality, $1 - \sqrt{5}$, is a negative value (since $\sqrt{5} \approx 2.236$ so $1 - \sqrt{5} \approx -1.236$). Therefore, while the mathematical solution to the inequality includes negative values, the physical meaning of $|z|$ restricts it to non-negative values. Substituting $|z|$ back: $1 - \sqrt{5} \le |z| \le 1 + \sqrt{5}$ Since $|z| \ge 0$ and $1 - \sqrt{5} < 0$, the effective lower bound for $|z|$ is 0. The valid range for $|z|$ is: $0 \le |z| \le 1 + \sqrt{5}$ From this range, it is clear that the maximum value of $|z|$ is $1 + \sqrt{5}$.

3. Common Mistakes & Tips

• Triangle Inequality Direction: Ensure you apply the triangle inequality correctly: $|z_1 + z_2| \le |z_1| + |z_2|$.
• Modulus Non-Negativity: Always remember that $|z| \ge 0$. This is crucial when interpreting the solution to the quadratic inequality.
• Algebraic Manipulation: Be careful with algebraic manipulations, especially when multiplying or dividing inequalities by variables. In this case, multiplying by $|z|$ was valid because we know $|z|$ is positive.

4. Summary

This problem utilizes the triangle inequality and the properties of complex number moduli to find the maximum possible value of $|z|$. By expressing $z$ in a convenient form, applying the triangle inequality, and solving the resulting quadratic inequality, we determine the range of possible values for $|z|$. Considering the non-negative nature of the modulus, we identify the maximum value as $1 + \sqrt{5}$.

5. Final Answer

The final answer is \boxed{\sqrt 5 + 1}, which corresponds to option (A).