Let the curve $z(1+i) + \overline{z}(1-i) = 4, z \in C$ divide the region $|z-3| \le 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha – \beta|$ equals:
- (1) $1 + \frac{\pi}{2}$
- (2) $1 + \frac{\pi}{3}$
- (3) $1 + \frac{\pi}{4}$
- (4) $1 + \frac{\pi}{6}$
Solution:
Let $z = x+iy$.
Equation: $(x+iy)(1+i) + (x-iy)(1-i) = 4$
$(x – y + i(x+y)) + (x – y – i(x+y)) = 4$
$2(x-y) = 4 \Rightarrow x – y = 2$.
Region $|z-3| \le 1$ is a circle centered at $(3,0)$ with radius $r=1$.
Distance of center $(3,0)$ from line $x-y-2=0$:
$d = \frac{|3 – 0 – 2|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.
Since $d < r$, the line cuts the circle.
At circle intersection: $(x-3)^2 + y^2 = 1$ and $y = x-2$.
$(x-3)^2 + (x-2)^2 = 1 \Rightarrow x^2-6x+9 + x^2-4x+4 = 1$
$2x^2 – 10x + 12 = 0 \Rightarrow x^2 – 5x + 6 = 0 \Rightarrow x=2, 3$.
Points: $(2,0)$ and $(3,1)$.
These points subtend a $90^\circ$ angle at the center $(3,0)$.
Area of smaller segment (Minor Area):
Area of sector ($90^\circ$) – Area of triangle.
$A_{minor} = \frac{1}{4}\pi(1)^2 – \frac{1}{2}(1)(1) = \frac{\pi}{4} – \frac{1}{2}$.
Total Area = $\pi(1)^2 = \pi$.
Major Area = $\pi – (\frac{\pi}{4} – \frac{1}{2}) = \frac{3\pi}{4} + \frac{1}{2}$.
Difference $|\alpha – \beta| = (\frac{3\pi}{4} + \frac{1}{2}) – (\frac{\pi}{4} – \frac{1}{2}) = \frac{2\pi}{4} + 1 = \frac{\pi}{2} + 1$.
Ans. (1)