Let the curve z(1+i)+bar{z}(1-i)=4, z in math | vedclass

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(A) Let $z=x+iy$. Substituting this into the equation of the curve:$(x+iy)(1+i)+(x-iy)(1-i)=4$$x+ix+iy-y+x-ix-iy-y=4$$2x-2y=4 \implies x-y=2$.The regi

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$1+\frac{\pi}{2}$

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(A) Let $z=x+iy$. Substituting this into the equation of the curve:$(x+iy)(1+i)+(x-iy)(1-i)=4$$x+ix+iy-y+x-ix-iy-y=4$$2x-2y=4 \implies x-y=2$.The region $|z-3| \leq 1$ represents a circle with center $(3,0)$ and radius $r=1$,given by $(x-3)^2+y^2 \leq 1$.The line $x-y=2$ intersects the circle at points where $(x-3)^2+(x-2)^2=1$,which simplifies to $x^2-6x+9+x^2-4x+4=1$,or $2x^2-10x+12=0$,so $x^2-5x+6=0$. The intersection points are $(2,0)$ and $(3,1)$.The distance from the center $(3,0)$ to the line $x-y-2=0$ is $d = \frac{|3-0-2|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.The area of the circular segment cut by the chord is $A_{segment} = r^2 \cos^{-1}(\frac{d}{r}) - d\sqrt{r^2-d^2} = 1^2 \cos^{-1}(\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{2}} = \frac{\pi}{4} - \frac{1}{2}$.Let $\alpha = \frac{\pi}{4} - \frac{1}{2}$ be the smaller area. The total area of the circle is $\pi r^2 = \pi$. The larger area is $\beta = \pi - (\frac{\pi}{4} - \frac{1}{2}) = \frac{3\pi}{4} + \frac{1}{2}$.Then $|\alpha-\beta| = |(\frac{\pi}{4} - \frac{1}{2}) - (\frac{3\pi}{4} + \frac{1}{2})| = |-\frac{2\pi}{4} - 1| = \frac{\pi}{2} + 1$.

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

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