1.Let the curve z(1+i)+zˉ(1−i)=4z(1 + i) + \bar{z}(1 - i) = 4z(1+i)+zˉ(1−i)=4, z∈Cz \in \mathbb{C}z∈C, divide the region ∣z−32∣≤12\left|z - \dfrac{3}{2}\right| \leq \dfrac{1}{2}z−23≤21 into two parts of areas α\alphaα and β\betaβ. Then α−β\alpha - \betaα−β equals:a.1+π21 + \dfrac{\pi}{2}1+2πb.1−π21 - \dfrac{\pi}{2}1−2πc.1+π41 + \dfrac{\pi}{4}1+4πd.1−π41 - \dfrac{\pi}{4}1−4πLogin to continueOnly logged in users canattempt or see the solution.
