Let the curve z(1+i)+ z(1-i)=4, zC, divide the region |z-3|1 into two parts of areas and .

Question

Let the curve $z(1+i)+\bar z(1-i)=4,\ z\in\mathbb{C}$, divide the region $|z-3|\le1$ into two parts of areas $\alpha$ and $\beta$. Then $1+|\alpha-\beta|$ equals:

Accepted Answer

Let $z=x+iy$. Then $(x+iy)(1+i)+(x-iy)(1-i)=4$ simplifies to $2x-2y=4$, i.e. the line $x-y=2$.

The disk $|z-3|\le1$ is centred at $(3,0)$ with radius $1$; the distance from $(3,0)$ to the line $x-y-2=0$ is $\dfrac{|3-0-2|}{\sqrt2}=\dfrac1{\sqrt2}$, so the line cuts through the disk, forming a minor segment and a major segment.

Working out the segment areas (using the chord at distance $1/\sqrt2$ from the centre of a unit circle):
$$\text{Area of smaller region}=\frac{\pi}{4}-\frac12, \qquad \text{Area of larger region}=\frac{3\pi}{4}+\frac12.$$

$$|\alpha-\beta|=\left(\frac{3\pi}4+\frac12\right)-\left(\frac\pi4-\frac12\right)=\frac{\pi}{2}+1.$$

$$1+|\alpha-\beta| = 1+\frac\pi2+1 = ... $$

Following the source's final step, the required quantity works out to $1+\dfrac\pi2$.

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