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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } z=x+i y \ & (x+i y)(1+i)+(x-i y)(1-i)=4 \ & x+i x+i y-y+x-i x-i y-y=4 \ & 2 x-2 y=4 \ & x-y=2 \ & |z-3| \leq 1 \ & (x-3)^2+y^2 \leq 1 \end{aligned}$ Area of shaded region $=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}$ Area of unshaded region inside the circle $=\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2}$ $	herefore$ difference of area $=\left(\frac{3 \pi}{4}+\frac{1}{2}ight)-\left(\frac{\pi}{4}-\frac{1}{2}ight)$ $=\frac{\pi}{2}+1$

Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in \mathrm{C}$, divide the region $|z-3| \leq 1$ into two parts of…

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