Let the curve $z(1+i)+\bar{z}(1-i)=4$, $z\in\mathbb{C}$, divide the region $|z-3|\leq1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals
Step-by-Step Solution
Key Concept: Convert the complex line to Cartesian form, compute the distance from the disc's centre to the line, and use the circular segment area formula to find $|\alpha-\beta|$.
Writing $z=x+iy$: $(x+iy)(1+i)+(x-iy)(1-i)=4 \Rightarrow 2x-2y=4 \Rightarrow x-y=2$.
Disc: centre $C=(3,0)$, radius $r=1$. Distance from $C$ to line $x-y-2=0$: $d=\dfrac{|3-0-2|}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}$.
**Minor segment** (smaller area, on the side of $C$ nearer the line):
$\alpha_\min = r^2\cos^{-1}\!\left(\dfrac{d}{r}\right)-d\sqrt{r^2-d^2} = \cos^{-1}\!\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{2}} = \dfrac{\pi}{4}-\dfrac{1}{2}$.
**Major segment:** $\pi r^2-\alpha_\min = \pi-\dfrac{\pi}{4}+\dfrac{1}{2}=\dfrac{3\pi}{4}+\dfrac{1}{2}$.
$|\alpha-\beta| = \left(\dfrac{3\pi}{4}+\dfrac{1}{2}\right)-\left(\dfrac{\pi}{4}-\dfrac{1}{2}\right) = \dfrac{\pi}{2}+1$.
Correct Answer: 1