<p>If \( \left| z - \dfrac{4}{z} \right| = 2 \), then the maximum value of \( |z| \) is:</p>
Step-by-Step Solution
Key Concept: Let r = |z|. By triangle inequality: ||z| - 4/|z|| \leq |z - 4/z| = 2. So |r - 4/r| \leq 2, giving r^2 - 2r - 4 \leq 0. Max r = 1 + \sqrt{5.}
<p>$ 2 = \left|z - \dfrac{4}{z}\right| \geq \bigl||z| - \dfrac{4}{|z|}\bigr| $. Let $r = |z|$: $|r - 4/r| \leq 2 \Rightarrow -2 \leq r - 4/r \leq 2 \Rightarrow r^2 - 2r - 4 \leq 0 \Rightarrow r \leq 1+\sqrt{5}$. Maximum is $1+\sqrt{5}$.</p>
Correct Answer: B