If z ≠ 0 be a complex number such that\(|z-\frac{1}{z}|=2\), then the maximum value of |z| is
- \(\sqrt2\)
- 1
- \(\sqrt2-1\)
- \(\sqrt2+1\)
The Correct Option is D
Solution and Explanation
\(|z-\frac{1}{z}|\)≥||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): \(\sqrt2+1\).
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.