Question

If \[ \left|z+\frac{2}{z}\right|=2, \] then the maximum value of \(|z|\) is:

• A. \(\sqrt{2}+1\)
• B. \(\sqrt{2}-1\)
• C. \(\sqrt{2}\)
• D. infinity

Hint:

For complex-number modulus questions involving \(z\) and \(\frac{1}{z}\), put \(z=re^{i\theta}\). This converts the expression into a condition involving \(r\) and \(\theta\).

Answer 1

The Correct Option is A

Step 1: Let \(|z|=r\).
Given, \[ \left|z+\frac{2}{z}\right|=2. \] Let \[ |z|=r,\qquad r>0. \] Using triangle inequality, \[ \left|z+\frac{2}{z}\right| \geq \left||z|-\left|\frac{2}{z}\right|\right|. \] Since \[ \left|\frac{2}{z}\right|=\frac{2}{|z|}=\frac{2}{r}, \] we get \[ 2\geq \left|r-\frac{2}{r}\right|. \]

Step 2: Solve the inequality.
\[ \left|r-\frac{2}{r}\right|\leq 2. \] So, \[ -2\leq r-\frac{2}{r}\leq 2. \] Since \(r>0\), multiply by \(r\): \[ -2r\leq r^2-2\leq 2r. \] From \[ r^2-2\leq 2r, \] we get \[ r^2-2r-2\leq 0. \] Solving, \[ r=1\pm \sqrt{3}. \] This gives the upper bound \[ r\leq 1+\sqrt{3}. \] But this is only a necessary bound from triangle inequality. We need the exact maximum.

Step 3: Use argument representation.
Let \[ z=re^{i\theta}. \] Then \[ \frac{2}{z}=\frac{2}{r}e^{-i\theta}. \] So, \[ z+\frac{2}{z} = re^{i\theta}+\frac{2}{r}e^{-i\theta}. \] Its modulus squared is \[ \left|z+\frac{2}{z}\right|^2 = r^2+\frac{4}{r^2}+4\cos 2\theta. \] Given modulus is \(2\), so \[ r^2+\frac{4}{r^2}+4\cos 2\theta=4. \]

Step 4: Find possible values of \(r\).
Since \[ -1\leq \cos 2\theta\leq 1, \] we need \[ -1\leq \frac{4-r^2-\frac{4}{r^2}}{4}\leq 1. \] The important condition for existence is \[ r^2+\frac{4}{r^2}-4\leq 4. \] Thus, \[ r^2+\frac{4}{r^2}\leq 8. \] Multiplying by \(r^2\), \[ r^4-8r^2+4\leq 0. \] Let \[ u=r^2. \] Then, \[ u^2-8u+4\leq 0. \] Solving, \[ u=4\pm 2\sqrt{3}. \] Thus, \[ r^2\leq 4+2\sqrt{3}. \] Therefore, \[ r\leq \sqrt{4+2\sqrt{3}}. \] Now, \[ 4+2\sqrt{3}=(\sqrt{3}+1)^2. \] So, \[ r\leq \sqrt{3}+1. \]

Step 5: Final conclusion.
Therefore, the maximum value of \(|z|\) is \[ \boxed{\sqrt{3}+1} \]