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

Concept:

Represent a complex number as $z = x + iy$.
A complex equation can be converted into a real equation representing a straight line or curve.
The region $|z-a| \le r$ represents a circle of radius $r$ centered at $a$.
Step 1: Let \[ z = x + iy \] Then, \[ z(1+i) + z(1-i) = (x+iy)(1+i) + (x+iy)(1-i) \]
Step 2: Simplify: \[ (x+iy)(1+i) = (x-y) + i(x+y) \] \[ (x+iy)(1-i) = (x+y) + i(y-x) \] Adding, \[ z(1+i) + z(1-i) = 2x + 2iy \] Given equation: \[ 2x + 2iy = 4 \Rightarrow x = 2 \] Thus, the curve is the straight line: \[ x = 2 \]
Step 3: Region $|z-3| \le 1$ represents the circle: \[ (x-3)^2 + y^2 \le 1 \] This is a circle of radius $1$ centered at $(3,0)$.
Step 4: The line $x=2$ cuts the circle, forming a chord. Distance of the line from center: \[ d = |3 - 2| = 1 \]
Step 5: Area of the circular segment: \[ \alpha = \frac{1}{2}r^2(\theta - \sin\theta) \] Here, \[ \cos\left(\frac{\theta}{2}\right) = \frac{d}{r} = 1 \Rightarrow \theta = \frac{\pi}{3} \]
Step 6: Area difference: \[ |\alpha - \beta| = \text{Area of circle} - 2\alpha \] \[ = \pi - \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) \] \[ |\alpha - \beta| = 1 + \frac{\pi}{6} \]