Let left|frac{bar{z}-i}{2 bar{z}+i}right|=fra | vedclass

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(A) Given $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$.Dividing numerator and denominator by $2$ inside the modulus: $\left|\frac{\bar{z}-

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$100$

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(A) Given $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$.Dividing numerator and denominator by $2$ inside the modulus: $\left|\frac{\bar{z}-i}{2(\bar{z}+i/2)}\right|=\frac{1}{3}$ $\Rightarrow \left|\frac{\bar{z}-i}{\bar{z}+i/2}\right|=\frac{2}{3}$.Let $z = x+iy$,then $\bar{z} = x-iy$. Substituting this:$3|x-iy-i| = 2|x-iy+i/2|$$9(x^2 + (-y-1)^2) = 4(x^2 + (-y+1/2)^2)$$9(x^2 + y^2 + 2y + 1) = 4(x^2 + y^2 - y + 1/4)$$9x^2 + 9y^2 + 18y + 9 = 4x^2 + 4y^2 - 4y + 1$$5x^2 + 5y^2 + 22y + 8 = 0$$x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0$.The center $C$ is $(0, -11/5)$.The area of the triangle with vertices $(0,0)$,$(0, -11/5)$,and $(\alpha, 0)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 11$.$\frac{1}{2} |0(-11/5 - 0) + 0(0 - 0) + \alpha(0 - (-11/5))| = 11$.$\frac{1}{2} |\alpha \cdot \frac{11}{5}| = 11$.$|\alpha| = 10$.Therefore,$\alpha^2 = 100$.

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$,where $z \in \mathbb{C}$,be the equation of a circle with center at $C$. If the area of the triangle,whose vertices are at the points $(0,0)$,$C$,and $(\alpha, 0)$,is $11$ square units,then $\alpha^2$ equals

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