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

Question

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

Accepted Answer

$$\begin{aligned}
& \text { Let } z=x+i y \
& (x+i y)(1+i)+(x-i y)(1-i)=4 \
& x+i x+i y-y+x-i x-i y-y=4 \
& 2 x-2 y=4 \
& x-y=2 \
& |z-3| \leq 1 \
& (x-3)^2+y^2 \leq 1
\end{aligned}$$

$$\begin{aligned}
&\text { Area of shaded region }=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}\
&\text { Area of unshaded region inside the circle }\
&\begin{aligned}
& =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \
& \therefore \text { difference of area }=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right)
\end{aligned}\
\&=\frac{\pi}{2}+1
\end{aligned}$$

Solutions · 1