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

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The curve is the line $x-y=2$. The centre $(3,0)$ is at distance $\frac1{\sqrt2}$ from it, so the chord subtends $\frac\pi2$ at the centre. Smaller segment $=\frac12\left(\frac\pi2-1\right)=\frac\pi4-\frac12$; larger $=\pi-\left(\frac\pi4-\frac12\right)$. Difference $=1+\frac\pi2$. Answer (1).

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

Explanation

Reviewed by Dr. Abubacker

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