Question:

Let \(S = z ∈ C: |z-3| <= 1\) and \(z (4+3i)+z(4-3)≤24.\)
If α + iβ is the point in S which is closest to 4i, then 25(α + β) is equal to ______.

Updated On: Mar 2, 2024
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Correct Answer: 80

Solution and Explanation

The correct answer is 80
Here |z – 3| < 1
\(⇒ (x – 3)^2 + y^2 < 1\)
And
\(z= (4+3i)+\overline{z}(4-3i)\) \(≤ 24\)
\(⇒ 4x-3y ≤ 12\)
\(tanθ = \frac{4}{3}\)
 

Figure.















∴ Coordinate of P = (3 – cosθ, sinθ)
\(=(3-\frac{3}{5},\frac{4}{5})\)
\(∴ α+iβ = \frac{12}{5}+\frac{4}{5}i\)
\(∴ 25(α+β)=80\)

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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.