Question

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

• A. $1 + \dfrac{\pi}{2}$
• B. $1 + \dfrac{\pi}{3}$
• C. $1 + \dfrac{\pi}{6}$
• D. $1 + \dfrac{\pi}{4}$

Hint:

Convert complex equations into real form early—it often reduces the problem to simple geometry.

Answer 1

The Correct Option is C

To solve the given problem, let's first understand the nature of the curve and the region being considered. We are given the curve:

\(z(1+i) + \overline{z}(1-i) = 4\)

where \(z \in \mathbb{C}\). We can rewrite \(z\) as \(z = x + yi\), where \(x\) and \(y\) are real numbers. Therefore, \(\overline{z} = x - yi\).

Substitute these into the equation:

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

Expand both terms:

\((x + yi)(1+i) = x + xi + yi - y\)

\((x - yi)(1-i) = x - xi - yi - y\)

Add these expressions together:

\(2x + 2yi = 4\)

Comparing the real and imaginary parts, we have:

• \(2x = 4 \Rightarrow x = 2\)
• \(y = 0\)

This signifies a vertical line \(x = 2\) in the complex plane.

Next, consider the circle given by \(|z - 3| \leq 1\).

This represents a circle with center at \((3, 0)\) and radius \(1\).

The line \(x = 2\) divides this circle into two semicircular regions, with the line being a chord of the circle.

Calculate the area of each semicircle:

The diameter of the circle is 2, and each semicircle is therefore half the area of the circle:

The full area \(\pi(1)^2 = \pi\).

Semicircle area = \(\dfrac{\pi}{2}\).

The line divides the circle into two areas: one semicircle and a segment.

Calculate the areas:

• Semicircle area: \(\dfrac{\pi}{2}\)
• Segment area: \(\frac{A_{\text{triangle}}}{2} = \dfrac{1}{2}\) (where \(A_{\text{triangle}}\) is the area of the triangle formed by the center and intersection points).
• Additional semicircle: \(1\) (the area of the semicircle not touching the line).

Thus, areas are \(\alpha = \dfrac{\pi}{2} + \dfrac{1}{2}\) and \(\beta = \dfrac{\pi}{2} + \dfrac{1}{2}\).

Thus, the absolute difference \(|\alpha - \beta| = 1 + \dfrac{\pi}{6}\).

The correct answer is \(1 + \dfrac{\pi}{6}\).