Question

If z ≠ 0 be a complex number such that\(|z-\frac{1}{z}|=2\), then the maximum value of |z| is

• A. \(\sqrt2\)
• B. 1
• C. \(\sqrt2-1\)
• D. \(\sqrt2+1\)

Answer 1

The Correct Option is D

To solve the given problem, we need to find the maximum value of \(|z|\) under the condition \(|z - \frac{1}{z}| = 2\).

Let \(z = re^{i\theta}\), where \(r = |z|\) and \(\theta\) is the argument of \(z\). The condition becomes:

\(\left|re^{i\theta} - \frac{1}{re^{i\theta}}\right| = 2\)

This simplifies to:

\(\left|re^{i\theta} - \frac{1}{r}e^{-i\theta}\right| = 2\)

\(\left|re^{i\theta} - \frac{1}{r}e^{-i\theta}\right| = \left|r\cos(\theta) + i r\sin(\theta) - \frac{1}{r}\cos(\theta) - i\frac{1}{r}\sin(\theta)\right|\)

Simplify it to:

\(\left|r - \frac{1}{r}\right|\left|\cos(\theta) + i\sin(\theta)\right| = 2\)

Since \(|\cos(\theta) + i\sin(\theta)| = 1\), we have:

\(\left|r - \frac{1}{r}\right| = 2\)

This results in two possible cases:

1. \(r - \frac{1}{r} = 2\)
2. \(r - \frac{1}{r} = -2\)

Case 1: \(r - \frac{1}{r} = 2\)

Here, we solve:

\(r^2 - 2r - 1 = 0\)

Using the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -2, c = -1\):

\(r = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-1)}}{2 \times 1}\)

\(r = \frac{2 \pm \sqrt{4 + 4}}{2}\)

\(r = \frac{2 \pm \sqrt{8}}{2}\)

\(r = \frac{2 \pm 2\sqrt{2}}{2}\)

\(r = 1 \pm \sqrt{2}\)

Since \(r > 0\), therefore \(r = 1 + \sqrt{2}\), which is approximately 2.414, the maximum value.

Case 2: \(r - \frac{1}{r} = -2\)

Here, we solve:

\(r^2 + 2r - 1 = 0\)

Using the quadratic formula:

\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 2, c = -1\):

\(r = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-1)}}{2 \times 1}\)

\(r = \frac{-2 \pm \sqrt{4 + 4}}{2}\)

\(r = \frac{-2 \pm \sqrt{8}}{2}\)

\(r = \frac{-2 \pm 2\sqrt{2}}{2}\)

\(r = -1 \pm \sqrt{2}\)

Since \(r > 0\), the viable solution here would give \(r = \sqrt{2} - 1\), which is invalid as maximum value of \(|z|\) must be greater.

Conclusively, the maximum value of \(|z|\) is \(\sqrt{2} + 1\).