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

We are given that \( |z - \frac{1}{z}| = 2 \) for a complex number \( z \neq 0 \). We need to find the maximum value of \( |z| \).

Let's analyze the given condition \( |z - \frac{1}{z}| = 2 \). 

Assume \( z = re^{i\theta} \) where \( r = |z| \) and \( \theta \) is the argument of \( z \). Thus, \( \frac{1}{z} = \frac{1}{r}e^{-i\theta} \). Then:

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

Simplifying, we have:

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

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

Using the properties of modulus (absolute value), we have:

\(\left( r - \frac{1}{r} \right)^2 \cos^2\theta + \left( r + \frac{1}{r} \right)^2 \sin^2\theta = |z - \frac{1}{z}|^2 = 4\)

We want this expression maximized, which means to focus on individual maximizers of both square terms. Notice:

\(\left( r - \frac{1}{r} \right)^2 + \left( r + \frac{1}{r} \right)^2 = |e^{2i\theta}|^2 ( \text{since } |e^{ix}| = 1 )\)

The expression maximizes when each part is in its maximum and when angles related to sine and cosine components fit functionality.

We equate terms for maximum attainable:

\(|r^2 - 1| = 2r\cos(\theta)\text{ » Max at r\cos\theta = 1}\)

Solving \( r^2 - 1 = 2r\cos\theta \), use quadratic formulae structure and trial solutions, which yield \( r = \sqrt{2} + 1 \) in tuning absolute max.

The maximal structure recognizes impact from roots, knowing quadratic form.

Therefore, the maximum value of \( |z| \) is \(\sqrt{2} + 1\), which concurs with option given.

Answer 2

\(|z-\frac{1}{z}|\)≥||z\(|\frac{-1}{z}|\)

⇒ \(||z|-\frac{1}{|z|}|\)≤2

Let |z| = r

\(|r-\frac{1}{r}|\)≤2

−2≤\(r-\frac{1}{r}\)≤2

\(r-\frac{1}{r}\)≥−2 and \(r-\frac{1}{r}\)≤2

\(r^2\)+2r–1≥0 and \(r^2\)–2r–1≤0

r∈[−∞,−1–\(\sqrt2\)]∪[−1+\(\sqrt2\),∞] and r∈[1−\(\sqrt2\), 1+\(\sqrt2\)]

Taking intersection r∈[\(\sqrt2-1,\sqrt2+1\)]
So, the correct option is (D): \(\sqrt2+1\).