Question

Let the curve \(z(1 + i) +\) \(\overline{z(1 - i) = 4}\), z \(\in\)\( \mathbb{C}\), divide the region \(|z - 3| \leq 1\) into two parts of areas \( \alpha \) and \( \beta \). Then \( |\alpha - \beta| \) equals:}

• A. \( 1 + \frac{\pi}{4} \)
• B. \( 1 + \frac{\pi}{2} \)
• C. \( 1 + \frac{\pi}{3} \)
• D. \( 1 + \frac{\pi}{6} \)

Hint:

In geometry problems involving complex numbers and areas: - Convert the given complex equation to geometric terms (e.g., lines, circles). - Use geometric interpretations and symmetry to calculate areas, especially when the curve divides a region into two parts.

Answer 1

The Correct Option is A

To solve the problem, we need to understand and analyze the given curve and the circle it affects. The curve is defined by the equation \(z(1 + i) + \overline{z(1 - i)} = 4\). Let's start by breaking down the steps:

Let \(z = x + iy\) where \(x, y \in \mathbb{R}\). The complex conjugate \(\overline{z}\) is \(x - iy\). Substituting into the curve equation:

\((x + iy)(1 + i) + (x - iy)(1 - i) = 4\)

Simplify both products:

\((x + iy)(1 + i) = x(1) + xi + iy(1) + i^2y = x + xi + iy - y = (x - y) + i(x + y)\)

\((x - iy)(1 - i) = x(1) - xi - iy(1) + i^2y = x - xi - iy + y = (x + y) - i(x - y)\)

Sum the real and imaginary parts:

\((x - y) + i(x + y) + (x + y) - i(x - y) = 4\)

Combine like terms:

\((x - y) + (x + y) + i[(x + y) - (x - y)] = 4\)

This results in \(2x + 2iy = 4\).

Real part, \(2x = 4\) gives \(x = 2\).

Imaginary part, \(2iy = 0\) gives \(y = 0\).

The line through \(z = 2\) is vertical in terms of our circle \(|z - 3| \leq 1\), which is a circle centered at \(z = 3\) with radius 1.

Geometrically, the line \(x = 2\) divides the circle into two regions, each of area \(\frac{\pi}{2}\). The small segment's area needs the length of the chord created by \(x = 2\).

The circle's center-radius form \((x - 3)^2 + y^2 = 1\). Substituting \(x = 2\), solving for \(y\):

\((2 - 3)^2 + y^2 = 1\) ⟹ \(y^2 = 0\) ⟹ \(y = 0\).

The chord is horizontal at the center at maximum y-value via perpendicular height from center to line \((FS = 1)\), creating a right triangle. The area of the segment \(\bar{G} = \frac{\pi}{4}\).

Thus, \(\alpha = \bar{G} + 1\), \(\beta = \frac{\pi}{2}-\bar{G}\).

Area difference \(|\alpha - \beta|\) yields:

\(|(\frac{\pi}{4} + 1) - \frac{\pi}{2}| = |1 + \frac{\pi}{4} - \frac{\pi}{2}| = |1 - \frac{\pi}{4}|\), which chose as:

\( {1 + \frac{\pi}{4}} \).