Step 1: Given Information
We are given that for any complex number \( w = c + id \), the argument \( \arg(w) \) lies in the range \( (-\pi, \pi] \), where \( i = \sqrt{-1} \). Also, we have real numbers \( \alpha \) and \( \beta \) such that for all complex numbers \( z = x + iy \), if \( \arg \left( \frac{z + \alpha}{z + \beta} \right) = \frac{\pi}{4} \), then the ordered pair \( (x, y) \) lies on the circle with equation:
\[
x^2 + y^2 + 5x - 3y + 4 = 0
\]
Our task is to find \( \alpha \) and \( \beta \) such that this condition holds.
Step 2: Geometrical Interpretation of the Problem
The condition \( \arg \left( \frac{z + \alpha}{z + \beta} \right) = \frac{\pi}{4} \) suggests a relationship between the points \( z = x + iy \), \( \alpha \), and \( \beta \) in the complex plane. The argument \( \frac{z + \alpha}{z + \beta} \) corresponds to the angle between the vectors \( z + \alpha \) and \( z + \beta \), and we are given that this angle is \( \frac{\pi}{4} \) (which is 45 degrees).
Step 3: Analyze the Circle Equation
The given equation of the circle is:
\[
x^2 + y^2 + 5x - 3y + 4 = 0
\]
We can complete the square for both \( x \) and \( y \) to rewrite this equation in standard form. Completing the square for \( x \) and \( y \), we get:
\[
(x + \frac{5}{2})^2 + (y - \frac{3}{2})^2 = \frac{25}{4} + \frac{9}{4} - 4
\]
Simplifying the constants:
\[
(x + \frac{5}{2})^2 + (y - \frac{3}{2})^2 = \frac{9}{4}
\]
So, the circle has center \( \left( -\frac{5}{2}, \frac{3}{2} \right) \) and radius \( \frac{3}{2} \).
Step 4: The Real Axis Intersections
The equation of the circle intersects the real axis (the \( x \)-axis) when \( y = 0 \). Substituting \( y = 0 \) into the equation of the circle, we get:
\[
x^2 + 5x + 4 = 0
\]
Solving this quadratic equation:
\[
x = \frac{-5 \pm \sqrt{5^2 - 4(1)(4)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2}
\]
The solutions are \( x = -4 \) and \( x = -1 \), so the circle intersects the real axis at \( (-4, 0) \) and \( (-1, 0) \).
Step 5: Condition on \( \alpha \) and \( \beta \)
The condition \( \arg \left( \frac{z + \alpha}{z + \beta} \right) = \frac{\pi}{4} \) suggests that the points \( \alpha \) and \( \beta \) are related geometrically to the center of the circle and the points on the real axis. After analyzing this geometrically, we conclude that the values of \( \alpha \) and \( \beta \) must satisfy \( \alpha = 1 \) and \( \beta = 4 \).
Step 6: Conclusion
The value of \( \alpha \beta = 1 \times 4 = 4 \).
Final Answer:
The value of \( \alpha \beta \) is \( \boxed{4} \).
If \( z \) is a complex number and \( k \in \mathbb{R} \), such that \( |z| = 1 \), \[ \frac{2 + k^2 z}{k + \overline{z}} = kz, \] then the maximum distance from \( k + i k^2 \) to the circle \( |z - (1 + 2i)| = 1 \) is:
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 