Question

If \(z\) is a complex number such that \[ \left|9z+\frac{1}{z}\right|=8, \] then the maximum value of \(|z|\) is

• A. \(\dfrac18\)
• B. \(\dfrac25\)
• C. \(\dfrac72\)
• D. \(1\)

Hint:

For expressions involving \[ \left|az+\frac{b}{z}\right|, \] let \[ |z|=r \] and use \[ \boxed{ \left|a+b\right| \ge \Big||a|-|b|\Big| } \] to obtain inequalities involving only \(r\).

Answer 1

The Correct Option is D

Step 1: Apply the triangle inequality. Using \[ |a+b|\ge\Big||a|-|b|\Big|, \] we get \[ 8 = \left|9z+\frac1z\right| \ge \left|9|z|-\frac1{|z|}\right|. \] Let \[ |z|=r>0. \] Then, \[ |9r-\tfrac1r|\le8. \]

Step 2: Solve the inequality. The inequality \[ -8\le 9r-\frac1r\le8 \] gives \[ -8r\le9r^2-1\le8r. \] From \[ 9r^2-8r-1\le0, \] we get \[ (9r+1)(r-1)\le0. \] Since \(r>0\), \[ r\le1. \] Also, \[ 9r^2+8r-1\ge0 \] gives \[ r\ge\frac19. \] Hence, \[ \boxed{\frac19\le r\le1.} \]

Step 3: Find the maximum value. Since \[ r=|z|, \] the maximum possible value is \[ \boxed{|z|_{\max}=1.} \] Hence, \[ \boxed{(D)} \] is the correct answer.