Question:
If $\left| z - \frac{3}{z} \right| = 2$, then the greatest value of $|z|$ is:
If $\left| z - \frac{3}{z} \right| = 2$, then the greatest value of $|z|$ is:
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For problems of the form $|z + k/z| = a$, the maximum value of $|z|$ is the positive root of the quadratic equation $r^2 - ar - |k| = 0$.
Updated On: May 2, 2026
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The Correct Option is C
Solution and Explanation
Concept:
Use the triangle inequality $|z_1| - |z_2| \leq |z_1 + z_2| \leq |z_1| + |z_2|$. Specifically, for $z_1 = z$ and $z_2 = -3/z$:
\[ | |z| - |3/z| | \leq | z - 3/z | \]
Step 1: Set up the inequality.
Given $|z - 3/z| = 2$: \[ |z| - \frac{3}{|z|} \leq 2 \]
Step 2: Solve the resulting quadratic inequality for $|z|$.
Let $r = |z|$. \[ r - \frac{3}{r} \leq 2 \quad \Rightarrow \quad r^2 - 2r - 3 \leq 0 \] Factorize: \[ (r - 3)(r + 1) \leq 0 \] The roots are $-1$ and $3$. Since $r$ must be positive, we look at the interval $0 < r \leq 3$.
Step 3: Determine the maximum value.
The inequality $r \leq 3$ indicates that the greatest possible value for the modulus $|z|$ is 3.
Step 1: Set up the inequality.
Given $|z - 3/z| = 2$: \[ |z| - \frac{3}{|z|} \leq 2 \]
Step 2: Solve the resulting quadratic inequality for $|z|$.
Let $r = |z|$. \[ r - \frac{3}{r} \leq 2 \quad \Rightarrow \quad r^2 - 2r - 3 \leq 0 \] Factorize: \[ (r - 3)(r + 1) \leq 0 \] The roots are $-1$ and $3$. Since $r$ must be positive, we look at the interval $0 < r \leq 3$.
Step 3: Determine the maximum value.
The inequality $r \leq 3$ indicates that the greatest possible value for the modulus $|z|$ is 3.
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