Let the curve $z(1+i)+\bar z(1-i)=4$, $z\in\mathbb{C}$, divide the region $|z-3|\le 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals:
Step-by-Step Solution
Key Concept: Writing $z=x+iy$, the line is $x-y=2$. The perpendicular distance from $(3,0)$ to this line is $1/\sqrt{2}$, so the chord subtends an angle of $\pi/2$ at the centre of the unit disk.
Let $z=x+iy$. Then $z(1+i)+\bar z(1-i)=(x+iy)(1+i)+(x-iy)(1-i)=2(x-y)$, so the curve is the line
$$L:\ x-y=2.$$
The disk $|z-3|\le 1$ has center $(3,0)$ and radius $1$. Distance from $(3,0)$ to $L$:
$$d=\frac{|3-0-2|}{\sqrt{2}}=\frac{1}{\sqrt{2}}.$$
The chord subtends a half-angle $\phi$ at the center with $\cos\phi=d/r=1/\sqrt{2}$, so $\phi=\pi/4$ and the full subtended angle is $\pi/2$.
\textbf{Smaller segment} (between chord and minor arc):
$$\alpha = \underbrace{\frac{1}{2}r^{2}\cdot \frac{\pi}{2}}_{\text{sector}}-\underbrace{\frac{1}{2}\cdot r^{2}\sin\frac{\pi}{2}}_{\text{triangle}}=\frac{\pi}{4}-\frac{1}{2}.$$
\textbf{Larger region} (rest of disk):
$$\beta = \pi r^{2}-\alpha = \pi - \left(\frac{\pi}{4}-\frac{1}{2}\right) = \frac{3\pi}{4}+\frac{1}{2}.$$
Hence $|\alpha-\beta| = \dfrac{3\pi}{4}+\dfrac{1}{2}-\dfrac{\pi}{4}+\dfrac{1}{2} = \dfrac{\pi}{2}+1.$
Correct Answer: 1