Let arg(z) represent the principal argument of the complex number z. Then, |z| = 3 and arg(z – 1) – arg(z + 1) = π/4 intersect
- exactly at one point
- exactly at two points
- nowhere
- at infinitely many points
The Correct Option is A
Solution and Explanation
The correct answer is (C) : nowhere
|z| = 3
arg(z-1) - arg(z+1) = π/4
\(∠AKL = ∠ACB = \frac{π}{4}\)
⇒ LK = AL = α = 1
K(0,1)
Radius = \(\sqrt2\)
PL = PK + KL = \(\sqrt2 + 1\)
P(0,1 + \(\sqrt2\))
Therefore , Number of intersection = 0
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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.