
$\begin{aligned}
& \text { Let } z=x+i y \
& (x+i y)(1+i)+(x-i y)(1-i)=4 \
& x+i x+i y-y+x-i x-i y-y=4 \
& 2 x-2 y=4 \
& x-y=2 \
& |z-3| \leq 1 \
& (x-3)^2+y^2 \leq 1
\end{aligned}Areaofshadedregion=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}Areaofunshadedregioninsidethecircle=\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2}\thereforedifferenceofarea=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{2}+1$