Question:medium

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

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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 \(\theta\).
Updated On: Jun 18, 2026
  • \(\sqrt{2}+1\)
  • \(\sqrt{2}-1\)
  • \(\sqrt{2}\)
  • infinity
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The Correct Option is A

Solution and Explanation

Step 1: Set up the modulus notation.
Let |z| = r with r>0. The given equation is |z + 2/z| = 2.

Step 2: Apply the reverse triangle inequality.

By the reverse triangle inequality, |z + 2/z| ≥ ||z| - |2/z|| = |r - 2/r|. Hence 2 ≥ |r - 2/r|, which is equivalent to |r - 2/r| ≤ 2.

Step 3: Convert to polar representation for an exact solution.

Write z = re^(iθ). Then 2/z = (2/r)e^(-iθ). Their sum is re^(iθ) + (2/r)e^(-iθ). The squared modulus is |z + 2/z|² = r² + 4/r² + 4cos 2θ. Setting this equal to 4 gives r² + 4/r² + 4cos 2θ = 4.

Step 4: Determine bounds on r using the range of cosine.

Since -1 ≤ cos 2θ ≤ 1, the equation requires r² + 4/r² - 4 ≤ 4, which simplifies to r² + 4/r² ≤ 8. Multiplying by r²: r⁴ - 8r² + 4 ≤ 0. Substitute u = r² to obtain u² - 8u + 4 ≤ 0. The quadratic roots are u = 4 ± 2√3. Thus r² ≤ 4 + 2√3, and taking square roots gives r ≤ √(4 + 2√3).

Step 5: Simplify the radical.

Recognize that 4 + 2√3 = (√3 + 1)², so r ≤ √3 + 1.

Step 6: Final conclusion.

The maximum possible value of |z| is √3 + 1.
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